You are given a graph with n nodes and m directed edges. One lowercase letter is assigned to each node. We define a path's value as the number of the most frequently occurring letter. For example, if letters on a path are "abaca", then the value of that path is 3. Your task is find a path whose value is the largest.

Input

The first line contains two positive integers n, m (1 ≤ n, m ≤ 300 000), denoting that the graph has n nodes and m directed edges.

The second line contains a string s with only lowercase English letters. The i-th character is the letter assigned to the i-th node.

Then m lines follow. Each line contains two integers x, y (1 ≤ x, y ≤ n), describing a directed edge from x to y. Note that x can be equal to y and there can be multiple edges between x and y. Also the graph can be not connected.

Output

Output a single line with a single integer denoting the largest value. If the value can be arbitrarily large, output -1 instead.

Examples

Input

5 4
abaca
1 2
1 3
3 4
4 5


Output

3


Input

6 6
xzyabc
1 2
3 1
2 3
5 4
4 3
6 4


Output

-1


Input

10 14
xzyzyzyzqx
1 2
2 4
3 5
4 5
2 6
6 8
6 5
2 10
3 9
10 9
4 6
1 10
2 8
3 7


Output

4

Note

In the first sample, the path with largest value is 1 → 3 → 4 → 5. The value is 3 because the letter 'a' appears 3 times.
#include <bits/stdc++.h>
using namespace std;
int fn(int N, int M, string S, int *x, int *y) {
  vector<vector<int>> adj(N);
  vector<int> inDeg(N);
  for (int i = 0; i < M; i++) {
    int x1, y1;
    x1 = x[i], y1 = y[i];
    x1--, y1--;
    adj[x1].push_back(y1);
    inDeg[y1]++;
  }
  vector<int> q;
  q.reserve(N);
  for (int i = 0; i < N; i++) {
    if (inDeg[i] == 0) q.push_back(i);
  }
  vector<array<int, 26>> dp(N, array<int, 26>({
                                   0,
                               }));
  int ans = 0;
  for (int z = 0; z < int(q.size()); z++) {
    int cur = q[z];
    ans = max(ans, ++dp[cur][S[cur] - 'a']);
    for (int nxt : adj[cur]) {
      for (int v = 0; v < 26; v++) {
        dp[nxt][v] = max(dp[nxt][v], dp[cur][v]);
      }
      if (!--inDeg[nxt]) {
        q.push_back(nxt);
      }
    }
  }
  if (q.size() >= N) {
    return ans;
  } else {
    return -1;
  }
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(nullptr);
  int N, M;
  cin >> N >> M;
  string S;
  cin >> S;
  int arr1[M], arr2[M];
  for (int i = 0; i < M; i++) {
    cin >> arr1[i] >> arr2[i];
  }
  cout << fn(N, M, S, arr1, arr2) << '\n';
  return 0;
}
