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;
char ch[300005];
int n, m, dp[300005][26];
bool in_stack[300005], vis[300005];
vector<vector<int> > adj;
bool check(int u) {
  if (in_stack[u])
    return true;
  else if (vis[u])
    return false;
  in_stack[u] = vis[u] = true;
  for (int v : adj[u]) {
    if (check(v)) return true;
  }
  in_stack[u] = false;
  return false;
}
int solve(int u, int c) {
  if (dp[u][c] == -1) {
    int ans = ((ch[u] - 'a') == c);
    int cv = ans;
    for (int v : adj[u]) ans = max(ans, cv + solve(v, c));
    dp[u][c] = ans;
  }
  return dp[u][c];
}
bool cycle() {
  for (int i = 0; i < n; i++)
    if (check(i)) return true;
  return false;
}
int main() {
  scanf("%d", &n), scanf("%d", &m);
  scanf("%s", ch);
  adj = vector<vector<int> >(n, vector<int>());
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d", &u), scanf("%d", &v);
    adj[u - 1].push_back(v - 1);
  }
  memset(dp, -1, sizeof(dp));
  memset(in_stack, 0, sizeof(in_stack));
  memset(vis, 0, sizeof(vis));
  if (cycle())
    printf("-1\n");
  else {
    int ans = 0;
    for (int i = 0; i < n; i++)
      for (int j = 0; j < 26; j++) ans = max(ans, solve(i, j));
    printf("%d\n", ans);
  }
  return 0;
}
