Problem: One 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 ai and bi (1 ≤ ai, bi ≤ 5;ai ≠ bi), where (ai, bi) 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

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;
public final class friends
{
    static FastScanner sc=new FastScanner(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out=new PrintWriter(System.out);
	static final String s1="WIN",s2="FAIL";
	
	public static void main(String args[]) throws Exception
	{
		int m=sc.nextInt(),max=-1;
		boolean[][] b1=new boolean[5][5];
		while(m>0)
		{
			int u=sc.nextInt()-1,v=sc.nextInt()-1;
			b1[u][v]=true;
			b1[v][u]=true;
			m--;
		}
		boolean ans=false;
		outer:
		for(int i=0;i<5;i++)
		{
			for(int j=0;j<5;j++)
			{
				if(i==j)
				{
					continue;
				}
				if(!b1[i][j])
				{
					for(int k=0;k<5;k++)
					{
						if(k==i || k==j)
						{
							continue;
						}
						if(!b1[i][k] && !b1[j][k])
						{
							ans=true;
							break outer;
						}
					}
				}
				else if(b1[i][j])
				{
					for(int k=0;k<5;k++)
					{
						if(k==i || k==j)
						{
							continue;
						}
						if(b1[i][k] && b1[j][k])
						{
							ans=true;
							break outer;
						}
					}
				}
			}
		}
		out.println((ans)?s1:s2);
		out.close();
	}
}
class FastScanner
{
    BufferedReader in;
    StringTokenizer st;

    public FastScanner(BufferedReader in) {
        this.in = in;
    }
	
    public String nextToken() throws Exception {
        while (st == null || !st.hasMoreTokens()) {
            st = new StringTokenizer(in.readLine());
        }
        return st.nextToken();
    }
	
	public String next() throws Exception {
		return nextToken().toString();
	}
	
    public int nextInt() throws Exception {
        return Integer.parseInt(nextToken());
    }

    public long nextLong() throws Exception {
        return Long.parseLong(nextToken());
    }

    public double nextDouble() throws Exception {
        return Double.parseDouble(nextToken());
    }
}