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.BufferedReader;
import java.io.PrintWriter;
import java.io.InputStreamReader;
import java.io.IOException;
import java.util.Collection;
import java.util.ArrayList;
import java.util.List;
import java.util.StringTokenizer;

public class CF76B implements Runnable{
        
    public static void main(String args[]){
       new CF76B().run();
    }
    
    @Override
    public void run(){
        try{
            in = new BufferedReader(new InputStreamReader(System.in));
            out = new PrintWriter(System.out);
            tok = null;
            solve();
            in.close();
            out.close();
            
        }
        catch(IOException e){
            e.printStackTrace();
            System.exit(0);
        }
    }
    
    int nextInt()throws IOException{
        return Integer.parseInt(nextToken());
    } 
    
    double nextDouble()throws IOException{
        return Double.parseDouble(nextToken());
    }
    
    long nextLong() throws IOException{
        return Long.parseLong(nextToken());
    }
    
    String nextToken()throws IOException{
        while(tok == null || !tok.hasMoreTokens()){
            tok = new StringTokenizer(in.readLine());
        }
        return tok.nextToken();
    }
    
    BufferedReader in;
    PrintWriter out;
    StringTokenizer tok;
    //////////////////////////////////////////////////////////////////
    //////////////////////////////////////////////////////////////////
    private void solve()throws IOException{
        int[][] adj = new int[5][5];
        for(int i = 0; i < 5; i++){
            adj[i][i] = -1;
        }
        int N = nextInt();
        for(int i = 0; i < N; i++){
            int a = nextInt() - 1;
            int b = nextInt() - 1;
            adj[a][b] = adj[b][a] = 1;
        }
        
        boolean acq = false;
        boolean nacq = false;
        
        for(int i = 0; i < 5; i++){
            int numa = 0;
            int numna = 0;
            for(int j = 0; j < 5; j++){
                if(adj[i][j] == 1)numa++;
                if(adj[i][j] == 0)numna++;
            }
            if(numa >= 3)acq = true;
            if(numna >= 3)nacq = true;
        }
        
        if(acq || nacq)out.println("WIN");
        else out.println("FAIL");
    }
}