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;
int dp[303030][30];
int u, v;
int in[303030];
vector<int> adj[303030];
char s[303030];
int main() {
  scanf("%d %d", &n, &m);
  scanf("%s", s);
  for (int i = 0; i < m; ++i) {
    scanf("%d %d", &u, &v);
    --u;
    --v;
    adj[u].push_back(v);
    in[v]++;
  }
  queue<int> q;
  for (int i = 0; i < n; ++i) {
    if (in[i] == 0) {
      q.push(i);
    }
  }
  int cnt = 0;
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    ++cnt;
    dp[u][s[u] - 'a']++;
    for (int v : adj[u]) {
      if (--in[v] == 0) {
        q.push(v);
      }
      for (int i = 0; i < 26; ++i) {
        dp[v][i] = max(dp[v][i], dp[u][i]);
      }
    }
  }
  if (cnt == n) {
    int ans = 0;
    for (int i = 0; i < n; ++i) {
      for (int j = 0; j < 26; ++j) {
        ans = max(ans, dp[i][j]);
      }
    }
    printf("%d\n", ans);
  } else {
    printf("-1\n");
  }
  return 0;
}
