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;
const int N = 300005;
int f[N][26];
int main() {
  int n, m;
  scanf("%d %d", &n, &m);
  string s;
  cin >> s;
  s = '*' + s;
  for (int i = 1; i <= n; i++) {
    f[i][s[i] - 'a']++;
  }
  vector<vector<int>> g(n + 1);
  vector<int> in_deg(n + 1, 0);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    g[x].push_back(y);
    in_deg[y]++;
  }
  queue<int> Q;
  for (int i = 1; i <= n; i++) {
    if (!in_deg[i]) {
      Q.push(i);
    }
  }
  int cnt = 0;
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (auto e : g[u]) {
      --in_deg[e];
      if (!in_deg[e]) {
        Q.push(e);
      }
      for (int i = 'a'; i <= 'z'; i++) {
        int a = i - 'a';
        f[e][a] = max(f[e][a], f[u][a] + (s[e] == i));
      }
    }
    cnt++;
  }
  if (cnt != n) {
    printf("-1");
    return 0;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 'a'; j <= 'z'; j++) {
      ans = max(ans, f[i][j - 'a']);
    }
  }
  printf("%d", ans);
  return 0;
}
