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;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long n, m, a, b, cnt = 0, ans = 0;
  string s;
  cin >> n >> m >> s;
  vector<long long> adj[n + 1], in(n + 1, 0);
  vector<vector<long long> > dp(n + 1, vector<long long>(26, 0));
  for (long long i = 0; i < m; i++) {
    cin >> a >> b;
    adj[a].push_back(b);
    in[b]++;
  }
  queue<long long> q;
  for (long long i = 1; i <= n; i++) {
    if (in[i] == 0) q.push(i), dp[i][s[i - 1] - 'a']++;
  }
  while (!q.empty()) {
    cnt++;
    long long v = q.front();
    q.pop();
    for (long long u : adj[v]) {
      for (long long i = 0; i < 26; i++)
        dp[u][i] = max(dp[u][i], dp[v][i] + (s[u - 1] - 'a' == i));
      in[u]--;
      if (in[u] == 0) q.push(u);
    }
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 0; j < 26; j++) ans = max(ans, dp[i][j]);
  }
  if (cnt != n) ans = -1;
  cout << ans;
}
