Lexing

Rupesh Nasre.

CS3300 Compiler Design

IIT Madras
August 2020
 Character stream

Machine-Independent
Machine-Independent
Lexical
LexicalAnalyzer
Analyzer Code
CodeOptimizer
Optimizer

Intermediate representation

Backend
Token stream
Frontend

Syntax
SyntaxAnalyzer
Analyzer Code
CodeGenerator
Generator

Syntax tree Target machine code

Machine-Dependent
Machine-Dependent
Semantic
SemanticAnalyzer
Analyzer Code
CodeOptimizer
Optimizer

Syntax tree Target machine code

Intermediate
Intermediate
Code Symbol
CodeGenerator
Generator Table
2
Intermediate representation
 Lexing Summary
Character stream
●
Basic lex Lexical
Machine-Indep.
Machine-Indep.
LexicalAnalyzer
Analyzer Code
CodeOptimizer
Optimizer
●
Input Buffering Token stream
Intermediate
representation

●
KMP String Matching Syntax
SyntaxAnalyzer
Analyzer Code
CodeGenerator
Generator

Syntax tree Target machine code

●
Regex → NFA → DFA Semantic
Machine-Dependent
Machine-Dependent
SemanticAnalyzer
Analyzer Code
CodeOptimizer
Optimizer
●
Regex → DFA Syntax tree Target machine code
Intermediate
Intermediate
Code
CodeGenerator
Generator
Intermediate
representation

3
 Role
●
Read input characters
●
Group into words (lexemes)
●
Return sequence of tokens
●
Sometimes
– Eat-up whitespace
– Remove comments
– Maintain line number information

4
 Token, Pattern, Lexeme
Token Pattern Sample lexeme
if Characters i, f if
comparison <= or >= or < or > or == or != <=, !=
identifier letter (letter + digit)* pi, score, D2
number Any numeric constant 3.14159, 0, 6.02e23
literal Anything but “, surrounded by “” “core dumped”

The following classes cover most or all of the tokens

●
One token for each keyword
●
Tokens for the operators, individually or in classes
●
Token for identifiers
●
One or more tokens for constants
●
One token each for punctuation symbols 5
 Representing Patterns
●
Keywords can be directly represented (break, int).
●
And so do punctuation symbols ({, +).
●
Others are finite, but too many!
– Numbers
– Identifiers
– They are better represented using a regular expression.
– [a-z][a-z0-9]*, [0-9]+

6
 Classwork: Regex Recap
●
If L is a set of letters (A-Z, a-z) and D is a set
of digits (0-9),
– Find the size of the language LD.
– Find the size of the language L U D.
– Find the size of the language L4.
●
Write regex for real numbers
– Without eE, without +- in mantissa (1.89)
– Without eE, with +- in mantissa (-1.89)
– With eE, with -+ in exponent (-1.89E-4)
7
 Classwork
●
Write regex for strings over alphabet {a, b} that
start and end with a.
●
Strings with third last letter as a.
●
Strings with exactly three bs.
●
Strings with even length.
●
Homework
– Exercises 3.3.6 from ALSU.

8
 Example Lex
/*/*variables
variables*/*/
Patterns [a-z]
[a-z] {{
yylval
yylval==*yytext
*yytext--'a';
'a';
return
returnVARIABLE;
VARIABLE; Tokens
}}

/*/*integers
integers*/*/
[0-9]+
[0-9]+ {{
yylval
yylval==atoi(yytext);
atoi(yytext); Lexemes
return
returnINTEGER;
INTEGER;
}}

/*/*operators
operators*/*/
[-+()=/*\n]
[-+()=/*\n]{{return
return*yytext;
*yytext;}}
/*/*skip
skipwhitespace
whitespace*/*/
[[\t]
\t] ;;

/*/*anything
anythingelse
elseis
isan
anerror
error*/*/
.. yyerror("invalid
yyerror("invalidcharacter");
character");
9
 a1.l a1.y

lex
lex yacc
yacc

lex.yy.c y.tab.c y.tab.h

gcc
gcc

Lexer and parser are not separate binaries;

This is your compiler. a.out they are part of the same executable.10
 Lex Regex
Expression Matches Example
c Character c a
\c Character c literally \*
“s” String s literally “**”
. Any character but newline a.*b
^ Beginning of a line ^abc
$ End of a line abc$
[s] Any of the characters in string s [abc]
[^s] Any one character not in string s [^abc]
r* Zero or more strings matching r a*
r+ One or more strings matching r a+
r? Zero or one r a?
r{m, n} Between m and n occurrences of r a{1,5}
r1r2 An r1 followed by an r2 ab
r1 | r2 An r1 or an r2 a|b
(r) Same as r (a | b) 11
r1/r2 r1 when followed by r2 abc/123
 Homework
●
Write a lexer to identify special words in a text.
– Words like stewardesses: only one hand
– Words like typewriter: only one keyboard row
– Words like skepticisms: alternate hands
●
Implement grep using lex with search pattern
as alphabetical text (no operators *, ?, ., etc.).

12
 Lexing and Context
●
Language design should ensure that lexing
can be done without context.
●
Your assignments and most languages need
context-insensitive lexing.

DO
DO55 I I==1.25
1.25 DO
DO55 I I==1,25
1,25

●
“DO 5 I” is an identifier in Fortran, as spaces are allowed in identifiers.
●
Thus, first is an assignment, while second is a loop.
●
Lexer doesn't know whether to consider the input “DO 5 I” as an identifier
or as a part of the loop, until parser informs it based on dot or comma.
●
Alternatively, lexer may employ a lookahead.
13
 Lexical Errors
●
It is often difficult to report errors for a lexer.
– fi (a == f(x)) ...
– A lexer doesn't know the context of fi. Hence it
cannot “see” the structure of the sentence –
structure is known only to the parser.
– fi = 2; OR fi(a == f(x));
●
But some errors a lexer can catch.
– 23 = @a;
– if $x friendof anil ...

What should a lexer do on catching an error? 14

 Error Handling
●
Multiple options
– exit(1);
– Panic mode recovery: delete enough input to recognize a
token
– Delete one character from the input
– Insert a missing character into the remaining input
– Replace a character by another character
– Transpose two adjacent characters
●
In practice, most lexical errors involve a single character.
●
Theoretical problem: Find the smallest number of
transformations (add, replace, delete) needed to convert the source
program into one that consists only of valid lexemes.
– Too expensive in practice to be worth the effort. 15
 Homework
●
Try exercise 3.1.2 from ALSU.

16
 Input Buffering
●
“We cannot know we were executing a finite
loop until we come out of the loop.”
●
In C, without reading the next character we
cannot determine a binary minus symbol (a-b).
 ->, -=, --, -e, ...
 Sometimes we may have to look several
characters in future, called lookahead.
 In the fortran example (DO 5 I), the lookahead
could be upto dot or comma.
●
Reading character-by-character from disk is
inefficient. Hence buffering is required. 17
 Input Buffering
●
A block of characters is read from disk into a buffer.
●
Lexer maintains two pointers:
– lexemeBegin
– forward E = M * C * * 2 \f

forward
lexemeBegin

What
Whatis
isthe
theproblem
problemwith
withsuch
suchaascheme?
scheme?
18
 Input Buffering
●
The issue arises when the lookahead is
beyond the buffer.
●
When you load the buffer, the previous content
is overwritten!
Input read Input to be
read

E = M * C * * 2 \f

forward
lexemeBegin

How
Howdo
dowe
wesolve
solvethis
thisproblem?
problem? 19
 Double Buffering
●
Uses two (half) buffers.
●
Assumes that the lookahead would not be
more than one buffer size.

Buf1 Buf2

E = M * C * * 2 \f

forward
lexemeBegin

20
 Transition Diagrams
●
Step to be taken on each character can be
specified as a state transition diagram.
– Sometimes, action may be associated with a state.
< =
0 1 2 return(comp, LE);
other yyless(1); return(comp, LT);
= 3

= return(comp, EQ);
4 5
>
other yyless(1); return(assign, ASSIGN);
6
= 8 return(comp, GE);
7
other 9 yyless(1); return(comp, GT);
21
...
 Keywords vs. Identifiers
●
Keywords may match identifier pattern
– Keywords: int, const, break, ...
– Identifiers: (alpha | _) (alpha | num | _)*
●
If unaddressed, may lead to strange errors.
– Install keywords a priori in the symbol table.
– Prioritize keywords
●
In lex, the rule for a keyword must precede
that of the identifier.

22

Incorrect (lex may give warning) Correct

 Special vs. General
●
In general, a specialized pattern must precede the
general pattern (associativity).
●
Lex also follows maximum substring matching rule
(precedence).
– Reordering the rules for < and <= would not affect the
functionality.
●
Compare with rule specialization in Prolog.
●
Classwork: Count number of he and she in a text.
●
Classwork: Write lex rules to recognize quoted
strings in C.
23
– Try to recognize \” inside it.
 he and she
she ++s; she {++s; REJECT;}
he ++h; he {++h;}
Retries another rule

What if I want to count all possible substrings he?

In general, the action associated with a rule may
not be easy / modular to duplicate.
Input: he ahe he she she fsfds fsf fs sfhe he she she she

he=5, she=5 he=10, she=5

24
 By the way...
●
Sometimes, you need not have a parser at all...
– You could define main in your lex file.
– Simply call yylex() from main.
– Compile using lex, then compile lex.yy.c using gcc
and execute a.out.

25
 Lookahead

Duniya usi ki hai jo aage dekhe

26
 Lookahead
●
Lexer needs to look into the future to know
where it is presently.

DO DO / .* COMMA { return DO;}

DO55 I I==1,25
1,25

●
/ signifies the lookahead symbol. The input is
read and matched, but is left unconsumed in
the current rule.

Corollary: DO loop index and increment must be on the same line

– no arbitrary whitespace allowed.
27
 String Matching
●
Lexical analyzer relies heavily on string
matching.
●
Given a program text T (length n) and a
pattern string s (length m), we want to check if
s occurs in T.
●
A naive algorithm would try all positions of T to
check for s (complexity O(m*n)).
n
T

m
s

28
 Where can we do better?
●
T = abababaababbbabbababb
●
s = ababaa

i=0
abababaababbbabbababb
ababaa

29
 Where can we do better?
●
T = abababaababbbabbababb
●
s = ababaa

i=0
abababaababbbabbababb
ababaa

30
 Where can we do better?
●
T = abababaababbbabbababb
●
s = ababaa

i=1
abababaababbbabbababb
ababaa

31
 Where can we do better?
●
T = abababaababbbabbababb
●
s = ababaa

i=2
abababaababbbabbababb
ababaa Match found

32
 Where can we do better?
●
T = abababaababbbabbababb
●
s = ababaa

T's current suffix

i=0
abababaababbbabbababb
ababaa
s's proper prefix

Key observation: T's current suffix which is a proper prefix in s

has the treasure for us.
Whenever there is a mismatch, we should utilize this overlap, 33
rather than restarting.
 Where can we do better?
●
T = abababaababbbabbababb
●
s = ababaa

T's current suffix

i=0
abababaababbbabbababb
ababaa
s's proper prefix

Key observation: T's current suffix which is a proper prefix in s

has the treasure for us.
Whenever there is a mismatch, we should utilize this overlap, 34
rather than restarting.
 Knuth-Morris-Pratt Algorithm
●
In 1970, Morris conceived the idea.
●
After a few weeks, Knuth independently discovered
the idea.
●
In 1970, Morris and Pratt published a techreport.
●
KMP published the algorithm jointly in 1977.
●
In 1969, Matiyasevic discovered a similar algorithm.

35
Source: wikipedia
 KMP String Matching
●
First linear time algorithm for string matching.
●
Whenver there is a mismatch, do not restart;
rather fail intelligently.
●
We define a failure function for each position,
taking into account the suffix and the prefix.
●
Note that the matched part of the large string T is
essentially the pattern string s. Thus, failure
function can be computed simply using pattern s.

abababaababbbabbababb
ababaa
36
 Failure is not final.

Failure function for ababaa

i 1 2 3 4 5 6
f(i) 0 0 1 2 3 1
seen a ab aba abab ababa ababaa
prefix ϵ ϵ a ab aba a

Algorithm given as Figure 3.19 in ALSU.

37
String matching with failure function
Text = a1a2...am; pattern = b1b2...bn (both indexed from 1)
s=0
for (i = 1; i <= m; ++i) { Go over Text
if (s > 0 && ai != bs+1) s = f(s) Handle failure
if (ai == bs+1) ++s Character match

if (s == n) return “yes” Full match

}
return “no”
i 1 2 3 4 5 6
f(i) 0 0 1 2 3 1
seen a ab aba abab ababa ababaa
prefix ϵ ϵ a ab aba a

38
String matching with failure function
Text = a1a2...am; pattern = b1b2...bn (both indexed from 1)
s=0
for (i = 1; i <= m; ++i) { Go over Text
while (s > 0 && ai != bs+1) s = f(s) Handle failure
if (ai == bs+1) ++s Character match

if (s == n) return “yes” Full match

}
return “no”
abababaababbbabbababb
ababaa

i 1 2 3 4 5 6
39
f(i) 0 0 1 2 3 1
 Classwork
●
Find failure function for pattern ababba.
●
Test it on string abababbaa.

●
Fibonacci strings are defined as
– s1 = b, s2 = a, sk = sk-1sk-2 for k > 2
– e.g., s3 = ab, s4 = aba, s5 = abaab
● Find the failure function for s6.

40
 Fibonacci Strings
– s1 = b, s2 = a, sk = sk-1sk-2 for k > 2
– e.g., s3 = ab, s4 = aba, s5 = abaab

●
Do not contain bb or aaa.
●
The words end in ba and ab alternatively.
●
Suppressing last two letters creates a palindrome.
●
...

Source: Wikipedia 41
 KMP Generalization
●
KMP can be used for keyword matching.
●
Aho and Corasick generalized KMP to
recognize any of a set of keywords in a text.
h e r s
0 1 2 8 9
i

s s
6 7

h e
3 4 5

Transition diagram for keywords he, she, his and hers.

i 1 2 3 4 5 6 7 8 9
f(i) 0 0 0 1 2 0 3 0 3 42
 KMP Generalization
●
When in state i, the failure function f(i) notes
the state corresponding to the longest proper
suffix that is also a prefix of some keyword.
h e r s
0 1 2 8 9
i

s s
6 7

h e
3 4 5

Transition diagram for keywords he, she, his and hers. In

Instate
state7,
7,character
character
ssmatches
matchesprefix
prefixofof
the keyword she
the keyword she to to
i 1 2 3 4 5 6 7 8 9
reach
reachstate
state3.
3.
f(i) 0 0 0 1 2 0 3 0 3 43
 Regex to DFA
●
Approach 1: Regex NFA DFA
●
Approach 2: Regex DFA
– The ideas would be helpful in parsing too.

44
 Regex NFA DFA
Draw an NFA for *cpp

Ʃ
c p p
0 1 2 3

p p
c
c p p
0 1 2 3
c c

How does a machine draw an NFA for an arbitrary

regular expression such as ((aa)*b(bb)*(aa)*)* ? 45
 Regex NFA DFA
●
For the sake of convenience, let's convert *cpp
into *abb and restrict to alphabet {a, b}.
●
Thus, the regex is (a|b)*abb.
●
How do we create an NFA for (a|b)*abb?
ϵ
a
ϵ ϵ
ϵ ϵ a b b
ϵ b ϵ

46
 Regex NFA DFA
●
For the sake of convenience, let's convert *cpp
into *abb and restrict to alphabet {a, b}.
●
Thus, the regex is (a|b)*abb.
●
How do we create an NFA for (a|b)*abb?
ϵ
a
ϵ 2 3 ϵ
0 ϵ 1 6 ϵ 7
a
8
b
9 b 10
ϵ 4 b ϵ
5

47
 Regex NFA DFA
NFA state DFA state a b
{0, 1, 2, 4, 7} A B C State
{1, 2, 3, 4, 6, 7, 8} B B D Transition
Table
{1, 2, 4, 5, 6, 7} C B C
{1, 2, 4, 5, 6, 7, 9} D B E
{1, 2, 4, 5, 6, 7, 10} E B C

ϵ
a
ϵ 2 3 ϵ
0 ϵ 1 6 ϵ 7
a
8
b
9 b 10
ϵ 4 b ϵ
5

48
 Regex NFA DFA
NFA state DFA state a b
{0, 1, 2, 4, 7} A B C State
{1, 2, 3, 4, 6, 7, 8} B B D Transition
Table
{1, 2, 4, 5, 6, 7} C B C
{1, 2, 4, 5, 6, 7, 9} D B E
{1, 2, 4, 5, 6, 7, 10} E B C

b
C
b a b
a b
A B D b E DFA
a a
a
49
Regex NFA DFA
Ʃ
a b b NFA
0 1 2 3

b b
a
a b b DFA
0 1 2 3
a a

b
C
b a b
a b
A B D b E DFA
a a non-minimal

a
50
 Regex NFA DFA
(a|b)*abb Regex

ϵ
a
ϵ 2 3 ϵ
0 ϵ 1 6 ϵ 7 a 8 b 9 b 10 NFA
ϵ b ϵ
4 5

ϵ
b
C
b a b
a b
A B D b E DFA
a a non-minimal

a
51
 Regex DFA
1. Construct a syntax tree for regex#.
2. Compute nullable, firstpos, lastpos, followpos.
3. Construct DFA using transition function.
4. Mark firstpos(root) as start state.
5. Mark states that contain position of # as
accepting states.

52
 Regex DFA
●
Regex is (a|b)*abb#.
●
Construct a syntax tree for the regex.
.

. #
. 6
b
. b 5
* 4
a
3
| ●
Leaves correspond to operands.
●
Interior nodes correspond to operators.
●
Operands constitute strings.

1 a b 2
53
 Functions from Syntax Tree
●
For a syntax tree node n
– nullable(n): true if n represents ϵ.
– firstpos(n): set of positions that correspond to the
first symbol of strings in n's subtree.
– lastpos(n): set of positions that correspond to the
last symbol of strings in n's subtree.
– followpos(n): set of next possible positions from n
for valid strings.
ϵ
a
ϵ 2 3 ϵ
0 ϵ 1 6 ϵ 7 a 8
b
9
b
10
ϵ 4 b 5 ϵ
54

ϵ
 nullable
●
nullable(n): true if n represents ϵ.
●
Regex is (a|b)*abb#.
F .

F . #
F . F
b
F . b F
T * F
a
F

F |

F a b F
55
 nullable
●
nullable(n): true if n represents ϵ.
Node n nullable(n)
leaf labeled ϵ true
leaf with position i false
or-node n = c1 | c2 nullable(c1) or nullable(c2)
cat-node n = c1c2 nullable(c1) and nullable(c2)
star-node n = c* true

Classwork: Write down the rules for firstpos(n).

●
firstpos(n): set of positions that correspond to the
first symbol of strings in n's subtree.
56
 firstpos
●
firstpos(n): set of positions that correspond
to the first symbol of strings in n's subtree.
Node n firstpos(n)
leaf labeled ϵ {}
leaf with position i {i}
or-node n = c1 | c2 firstpos(c1) U firstpos(c2)
cat-node n = c1c2
star-node n = c* firstpos(c)

57
 firstpos
●
firstpos(n): set of positions that correspond
to the first symbol of strings in n's subtree.
Node n firstpos(n)
leaf labeled ϵ {}
leaf with position i {i}
or-node n = c1 | c2 firstpos(c1) U firstpos(c2)
cat-node n = c1c2 if (nullable(c1)) firstpos(c1) U firstpos(c2)
else firstpos(c1)

star-node n = c* firstpos(c)

Classwork: Write down the rules for lastpos(n).

58
 lastpos
●
lastpos(n): set of positions that correspond
to the last symbol of strings in n's subtree.
Node n lastpos(n)
leaf labeled ϵ {}
leaf with position i {i}
or-node n = c1 | c2 lastpos(c1) U lastpos(c2)
cat-node n = c1c2 if (nullable(c2)) lastpos(c1) U lastpos(c2)
else lastpos(c2)

star-node n = c* lastpos(c)

59
 firstpos lastpos
{1,2,3} {6} .

{1,2,3} {5} .
#
{1,2,3} {4} . 6
b {6} {6}
{1,2,3} {3} . b 5
{5} {5}
{1,2} {1,2} * 4
a {4} {4}
3
{3} {3}
{1,2} {1,2} |

1 a b 2
{1} {1} {2} {2}

60
 followpos
●
followpos(n): set of next possible positions
from n for valid strings.
– If n is a cat-node with child nodes c1 and c2, then
for each position in lastpos(c1), all positions in
firstpos(c2) follow.
– If n is a star-node, then for each position in
lastpos(n), all positions in firstpos(n) follow.

61
 followpos
If n is a cat-node with child nodes c1 and c2, then for each position in
lastpos(c1), all positions in firstpos(c2) follow.
{1,2,3} {6} .

{1,2,3} {5} .
#
{1,2,3} {4} . 6
b {6} {6}
{1,2,3} {3} . b 5
{5} {5}
{1,2} {1,2} * 4
a {4} {4}
3 n followpos(n)
{3} {3}
{1,2} {1,2} | 1 {3}
2 {3}

1 a b 2
{1} {1} {2} {2}
62
 followpos
If n is a cat-node with child nodes c1 and c2, then for each position in
lastpos(c1), all positions in firstpos(c2) follow.
{1,2,3} {6} .

{1,2,3} {5} .
#
{1,2,3} {4} . 6
b {6} {6}
{1,2,3} {3} . b 5
{5} {5}
{1,2} {1,2} * 4
a {4} {4}
3 n followpos(n)
{3} {3}
{1,2} {1,2} | 1 {3}
2 {3}
3 {4}
1 a b 2 4 {5}
{1} {1} {2} {2} 5 {6}
6 {} 63
 followpos
If n is a star-node, then for each position in lastpos(n), all positions in
firstpos(n) follow.

{1,2,3} {6} .

{1,2,3} {5} .
#
{1,2,3} {4} . 6
b {6} {6}
{1,2,3} {3} . b 5
{5} {5}
{1,2} {1,2} * 4
a {4} {4}
3 n followpos(n)
{3} {3}
{1,2} {1,2} | 1 {3}
2 {3}
3 {4}
1 a b 2 4 {5}
{1} {1} {2} {2} 5 {6}
6 {} 64
 followpos
If n is a star-node, then for each position in lastpos(n), all positions in
firstpos(n) follow.

{1,2,3} {6} .

{1,2,3} {5} .
#
{1,2,3} {4} . 6
b {6} {6}
{1,2,3} {3} . b 5
{5} {5}
{1,2} {1,2} * 4
a {4} {4}
3 n followpos(n)
{3} {3}
{1,2} {1,2} | 1 {3, 1, 2}
2 {3, 1, 2}
3 {4}
1 a b 2 4 {5}
{1} {1} {2} {2} 5 {6}
6 {} 65
 Regex DFA
1.Construct a syntax tree for regex#.
2.Compute nullable, firstpos, lastpos, followpos.
3.Construct DFA using transition function (next slide).
4.Mark firstpos(root) as start state.
5.Mark states that contain position of # as
accepting states.

66
 DFA Transitions
create unmarked state firstpos(root). {1,2,3} {6} .
while there exists unmarked state s {
mark s a b a
1 2 3
for each input symbol x {
uf = U followpos(p) where p is in s labeled x
transition[s, x] = uf n followpos(n)
1 {3, 1, 2}
if uf is newly created
2 {3, 1, 2}
unmark uf b 3 {4}
a
} 123 1234 4 {5}
5 {6} 67
} 6 {}
 Final DFA
b

b a
a b b DFA
123 1234 1235 1236
a
a

Ʃ
a b b NFA
0 1 2 3

b b
a
a b b DFA
0 1 2 3
a a
68
 Regex DFA
1.Construct a syntax tree for regex#.
2.Compute nullable, firstpos, lastpos, followpos.
3.Construct DFA using transition function.
4.Mark firstpos(root) as start state.
5.Mark states that contain position of # as
accepting states.

Do this for (b|ab)*(aa|b)*.

69
 In case you are wondering...
●
What to do with this DFA?
– Recognize strings during lexical analysis.
– Could be used in utilities such as grep.
– Could be used in regex libraries as supported in
php, python, perl, ....

70
 Lexing Summary
Character stream
●
Basic lex Lexical
Machine-Indep.
Machine-Indep.
LexicalAnalyzer
Analyzer Code
CodeOptimizer
Optimizer
●
Input Buffering Token stream Intermediate
representation

●
KMP String Matching Syntax
SyntaxAnalyzer
Analyzer Code
CodeGenerator
Generator

Syntax tree Target machine code

●
Regex → NFA → DFA Semantic
Machine-Dependent
Machine-Dependent
SemanticAnalyzer
Analyzer Code
CodeOptimizer
Optimizer
●
Regex → DFA Syntax tree Target machine code
Intermediate
Intermediate
Code
CodeGenerator
Generator
Intermediate
representation

71