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;
long long degree[300005];
long long dp[300005][26];
string str;
queue<long long> q;
int main() {
  long long n, m;
  cin >> n >> m;
  cin >> str;
  vector<long long> *adjlist;
  adjlist = new vector<long long>[n + 1];
  for (long long r = 1; r <= n; r++) {
    degree[r] = 0;
  }
  for (long long r = 0; r < m; r++) {
    long long src, des;
    cin >> src >> des;
    adjlist[src].push_back(des);
    degree[des]++;
  }
  long long cnt = 0;
  for (long long r = 1; r <= n; r++) {
    if (degree[r] == 0) {
      q.push(r);
    }
  }
  for (long long r = 1; r <= n; r++) {
    for (long long s = 0; s < 26; s++) {
      dp[r][s] = 0;
    }
  }
  long long ans = 1;
  while (!q.empty() && cnt <= n) {
    long long f = q.front();
    q.pop();
    long long num = int(str[f - 1]) - 97;
    dp[f][num]++;
    ans = max(ans, dp[f][num]);
    for (long long r = 0; r < adjlist[f].size(); r++) {
      long long neigh = adjlist[f][r];
      for (long long s = 0; s < 26; s++) {
        dp[neigh][s] = max(dp[f][s], dp[neigh][s]);
        ans = max(ans, dp[neigh][s]);
      }
      degree[neigh]--;
      if (degree[neigh] == 0) {
        q.push(neigh);
      }
    }
    cnt++;
  }
  while (!q.empty()) {
    q.pop();
  }
  if (cnt == n) {
    cout << ans;
    return 0;
  }
  cout << "-1";
  return 0;
}
