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 power(long long a, long long b) {
  long long ans = 1;
  while (b > 0) {
    if (b % 2 != 0) {
      ans = (ans * a) % 1000000007;
    }
    a = (a * a) % 1000000007;
    b >>= 1;
  }
  return ans;
}
int o = 1;
void dfs(vector<int> v[], int a[][26], string& s, bool vis[], bool cur[],
         int i) {
  if (o == 0) return;
  cur[i] = true;
  a[i][s[i] - 'a']++;
  int b[26];
  memset(b, 0, sizeof(b));
  for (auto j : v[i]) {
    if (cur[j] == true) {
      o = 0;
      break;
    }
    if (vis[j] == false) {
      dfs(v, a, s, vis, cur, j);
    }
    for (int k = 0; k < 26; k++) {
      b[k] = max(b[k], a[j][k]);
    }
  }
  for (int k = 0; k < 26; k++) a[i][k] += b[k];
  cur[i] = false;
  vis[i] = true;
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t = 1;
  while (t--) {
    int n, m, p, q;
    string s;
    cin >> n >> m >> s;
    s = '.' + s;
    vector<int> v[n + 1];
    int a[n + 1];
    memset(a, 0, sizeof(a));
    for (int i = 0; i < m; i++) {
      cin >> p >> q;
      if (p == q) {
        cout << -1;
        return 0;
      }
      v[p].push_back(q);
      a[q]++;
    }
    int b[n + 1][26];
    int k = 0;
    bool vis[n + 1], cur[n + 1];
    memset(vis, false, sizeof(vis));
    memset(cur, false, sizeof(cur));
    memset(b, 0, sizeof(b));
    for (int i = 1; i <= n; i++) {
      if (o == 0) break;
      if (a[i] == 0) {
        k = 1;
        dfs(v, b, s, vis, cur, i);
      }
    }
    int ma = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < 26; j++) {
        ma = max(ma, b[i][j]);
      }
    }
    if (k == 0 || o == 0) {
      cout << -1;
      return 0;
    }
    cout << ma;
  }
  return 0;
}
