Virtual contest is a way to take part in past contest, as close as possible to participation on time. It is supported only ACM-ICPC mode for virtual contests.
If you've seen these problems, a virtual contest is not for you - solve these problems in the archive.
If you just want to solve some problem from a contest, a virtual contest is not for you - solve this problem in the archive.
Never use someone else's code, read the tutorials or communicate with other person during a virtual contest.

No tag edit access

B. Friends

time limit per test

1 secondmemory limit per test

256 megabytesinput

standard inputoutput

standard outputOne day Igor K. stopped programming and took up math. One late autumn evening he was sitting at a table reading a book and thinking about something.

The following statement caught his attention: "Among any six people there are either three pairwise acquainted people or three pairwise unacquainted people"

Igor just couldn't get why the required minimum is 6 people. "Well, that's the same for five people, too!" — he kept on repeating in his mind. — "Let's take, say, Max, Ilya, Vova — here, they all know each other! And now let's add Dima and Oleg to Vova — none of them is acquainted with each other! Now, that math is just rubbish!"

Igor K. took 5 friends of his and wrote down who of them is friends with whom. Now he wants to check whether it is true for the five people that among them there are either three pairwise acquainted or three pairwise not acquainted people.

Input

The first line contains an integer *m* (0 ≤ *m* ≤ 10), which is the number of relations of acquaintances among the five friends of Igor's.

Each of the following *m* lines contains two integers *a*_{i} and *b*_{i} (1 ≤ *a*_{i}, *b*_{i} ≤ 5;*a*_{i} ≠ *b*_{i}), where (*a*_{i}, *b*_{i}) is a pair of acquainted people. It is guaranteed that each pair of the acquaintances is described exactly once. The acquaintance relation is symmetrical, i.e. if *x* is acquainted with *y*, then *y* is also acquainted with *x*.

Output

Print "FAIL", if among those five people there are no either three pairwise acquainted or three pairwise unacquainted people. Otherwise print "WIN".

Examples

Input

4

1 3

2 3

1 4

5 3

Output

WIN

Input

5

1 2

2 3

3 4

4 5

5 1

Output

FAIL

Codeforces (c) Copyright 2010-2018 Mike Mirzayanov

The only programming contests Web 2.0 platform

Server time: Feb/16/2019 13:41:58 (d3).

Desktop version, switch to mobile version.

Supported by

User lists

Name |
---|