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 n, m;
string s;
char nd[300005];
vector<int> adj[300005];
int deg[300005];
int dp[300005][26];
stack<int> st;
int main() {
  scanf("%d%d", &n, &m);
  cin >> s;
  for (int i = (0); i < (s.length()); ++i) nd[i + 1] = s[i];
  for (int i = (0); i < (m); ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    adj[u].push_back(v);
    deg[v]++;
  }
  int cnt = 0;
  for (int i = (0); i < (n); ++i) {
    dp[i + 1][nd[i + 1] - 'a'] = 1;
    if (!deg[i + 1]) {
      st.push(i + 1);
      cnt++;
    }
  }
  while (!st.empty()) {
    int cn = st.top();
    st.pop();
    for (int i = (0); i < (adj[cn].size()); ++i) {
      int cadj = adj[cn][i];
      if (--deg[cadj] == 0) {
        st.push(cadj);
        cnt++;
      }
      for (int i = 0; i < 26; i++)
        dp[cadj][i] = (dp[cadj][i] >= dp[cn][i] ? dp[cadj][i] : dp[cn][i]);
      dp[cadj][nd[cadj] - 'a'] =
          (dp[cadj][nd[cadj] - 'a'] >= dp[cn][nd[cadj] - 'a'] + 1
               ? dp[cadj][nd[cadj] - 'a']
               : dp[cn][nd[cadj] - 'a'] + 1);
    }
  }
  if (cnt != n) {
    cout << -1 << endl;
  } else {
    int ans = 0;
    for (int i = (0); i < (n); ++i)
      for (int j = (0); j < (26); ++j)
        ans = (ans >= dp[i + 1][j] ? ans : dp[i + 1][j]);
    cout << ans << endl;
  }
  return 0;
}
