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, a, b, dp[300005][27], branch[300005][27], vis[300005];
char s[300005];
vector<int> v[300005];
bool findcycle(int node) {
  vis[node] = 1;
  for (int child : v[node]) {
    if (vis[child] == 1) return true;
    if (vis[child] == -1) continue;
    if (findcycle(child) == true) return true;
  }
  vis[node] = -1;
  return false;
}
void dfs(int node) {
  int flag = 0;
  vis[node] = 1;
  for (int child : v[node]) {
    flag = 0;
    if (vis[child] == 0) dfs(child);
    for (int j = 0; j < 26; j++) {
      branch[node][j] = max(branch[node][j], branch[child][j]);
    }
  }
  branch[node][s[node] - 'a']++;
}
int main() {
  int ans = 0;
  scanf("%d%d", &n, &m);
  cin >> s + 1;
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    v[a].push_back(b);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      if (findcycle(i) == true) {
        printf("-1\n");
        return 0;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    vis[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs(i);
    for (int j = 0; j < 26; j++) {
      ans = max(ans, max(branch[i][j], dp[i][j]));
    }
  }
  printf("%d\n", ans);
  return 0;
}
