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;
struct Edge {
  int from, to, next;
} edge[300010];
int head[300010], tot;
inline void addedge(int u, int v) {
  edge[++tot] = (Edge){u, v, head[u]}, head[u] = tot;
}
int f[300010][26], n, m, d[300010];
char s[300010];
queue<int> Q;
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    addedge(x, y);
    d[y]++;
  }
  for (int i = 1; i <= n; i++) {
    if (!d[i]) {
      Q.push(i);
      f[i][s[i] - 'a'] = 1;
    }
  }
  int rem = n;
  while (!Q.empty()) {
    int now = Q.front();
    Q.pop();
    rem--;
    for (int i = head[now]; i; i = edge[i].next) {
      Edge e = edge[i];
      for (int j = 0; j < 26; j++) {
        f[e.to][j] = max(f[e.to][j], f[now][j] + (s[e.to] - 'a' == j));
      }
      d[e.to]--;
      if (!d[e.to]) Q.push(e.to);
    }
  }
  if (rem) return puts("-1"), 0;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 26; j++) {
      ans = max(ans, f[i][j]);
    }
  }
  printf("%d\n", ans);
  return 0;
}
