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 CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9 + 7;
const long long n_ = 1e5 + 1000;
const long double PI = acos(-1.0);
long long gcd(long long a, long long b) { return (a ? gcd(b % a, a) : b); }
long long power(long long a, long long n) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
    }
    n >>= 1;
    a *= a;
  }
  return p;
}
long long power(long long a, long long n, long long mod) {
  long long p = 1;
  while (n > 0) {
    if (n % 2) {
      p = p * a;
      p %= mod;
    }
    n >>= 1;
    a *= a;
    a %= mod;
  }
  return p % mod;
}
bool vowel(char ch) {
  if (ch == 'a' || ch == 'e' || ch == 'i' || ch == 'o' || ch == 'u')
    return true;
  else
    return false;
}
class UnionFind {
 private:
  vector<long long> p, rank, setSize;
  long long numSets;

 public:
  UnionFind(long long N) {
    setSize.assign(N, 1);
    numSets = N;
    rank.assign(N, 0);
    p.assign(N, 0);
    for (long long i = 0; i < N; i++) p[i] = i;
  }
  long long findSet(long long i) {
    return (p[i] == i) ? i : (p[i] = findSet(p[i]));
  }
  bool isSameSet(long long i, long long j) { return findSet(i) == findSet(j); }
  void unionSet(long long i, long long j) {
    if (!isSameSet(i, j)) {
      numSets--;
      long long x = findSet(i), y = findSet(j);
      if (rank[x] > rank[y]) {
        p[y] = x;
        setSize[x] += setSize[y];
      } else {
        p[x] = y;
        setSize[y] += setSize[x];
        if (rank[x] == rank[y]) rank[y]++;
      }
    }
  }
  long long numDisjointSets() { return numSets; }
  long long sizeOfSet(long long i) { return setSize[findSet(i)]; }
};
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long n, a, b;
  cin >> n;
  long long arr[6][6];
  memset(arr, 0, sizeof(arr));
  for (long long i = 0; i < n; i++) {
    cin >> a >> b;
    arr[a][b] = 1;
    arr[b][a] = 1;
  }
  for (long long i = 1; i <= 5; i++) {
    for (long long j = i + 1; j <= 5; j++) {
      for (long long k = j + 1; k <= 5; k++) {
        if (arr[i][j] == arr[i][k] && arr[i][j] == arr[j][k]) {
          cout << "WIN";
          return 0;
        }
      }
    }
  }
  cout << "FAIL";
  return 0;
}