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 maxn = 300000 + 100;
const int maxm = 600000 + 100;
int head[maxn], to[maxm], Next[maxm], in[maxn];
int f[maxn][30];
char w[maxn];
int n, m, sz;
void add_edge(int a, int b) {
  ++sz;
  to[sz] = b;
  Next[sz] = head[a];
  head[a] = sz;
}
int main() {
  scanf("%d%d", &n, &m);
  for (int i = 1; i <= n; i++) scanf(" %c", &w[i]);
  int a, b;
  sz = 0;
  memset(head, -1, sizeof(head));
  memset(in, 0, sizeof(in));
  for (int i = 1; i <= m; i++) {
    scanf("%d%d", &a, &b);
    in[b]++;
    add_edge(a, b);
  }
  memset(f, 0, sizeof(f));
  queue<int> Q;
  for (int i = 1; i <= n; i++) {
    if (!in[i]) {
      Q.push(i);
      f[i][w[i] - 'a'] = 1;
    }
  }
  while (!Q.empty()) {
    int u = Q.front();
    Q.pop();
    for (int i = head[u]; i != -1; i = Next[i]) {
      int v = to[i];
      --in[v];
      for (int j = 0; j < 26; j++) {
        f[v][j] = max(f[v][j], f[u][j]);
      }
      if (!in[v]) {
        Q.push(v);
        f[v][w[v] - 'a']++;
      }
    }
  }
  for (int i = 1; i <= n; i++) {
    if (in[i]) {
      printf("-1");
      return 0;
    }
  }
  int ans = -1;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 26; j++) {
      ans = max(ans, f[i][j]);
    }
  }
  printf("%d", ans);
  return 0;
}
