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;
void maximiza(int &a, int b) { a = max(a, b); }
const int limite = 500000;
int n, m;
vector<int> g[limite];
vector<int> ginv[limite];
int maximo[limite][26];
int valor[limite];
int aridad[limite];
queue<int> q;
int sol;
int main() {
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    char c;
    cin >> c;
    valor[i] = c - 'a';
  }
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    ginv[v].push_back(u);
    aridad[u]++;
  }
  for (int u = 1; u <= n; u++) {
    if (aridad[u] == 0) {
      q.push(u);
      maximo[u][valor[u]] = 1;
    }
  }
  while (not q.empty()) {
    int u = q.front();
    q.pop();
    {
      vector<int> &ar = g[u];
      for (int i = 0; i < int(ar.size()); i++) {
        int v = ar[i];
        for (int c = 0; c < 26; c++)
          maximiza(maximo[u][c], maximo[v][c] + (c == valor[u]));
      }
    }
    {
      vector<int> &ar = ginv[u];
      for (int i = 0; i < int(ar.size()); i++) {
        int v = ar[i];
        aridad[v]--;
        if (aridad[v] == 0) q.push(v);
      }
    }
  }
  for (int u = 1; u <= n; u++) {
    if (aridad[u]) {
      cout << -1 << endl;
      exit(0);
    }
  }
  for (int u = 1; u <= n; u++)
    for (int c = 0; c < 26; c++) maximiza(sol, maximo[u][c]);
  cout << sol << endl;
}
