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, id[300000];
char s[300001];
vector<int> adj[300000];
int memo[300000][26];
int solve(int n, char c) {
  if (memo[n][c - 'a'] > -1) return memo[n][c - 'a'];
  int h = 0;
  for (int v : adj[n]) h = max(h, solve(v, c));
  return memo[n][c - 'a'] = h + (c == s[n]);
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s);
  memset(id, 0, sizeof id);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    x--;
    y--;
    adj[x].push_back(y);
    id[y]++;
  }
  queue<int> que;
  for (int i = 0; i < n; i++)
    if (id[i] == 0) que.push(i);
  int taken = 0;
  while (!que.empty()) {
    int t = que.front();
    que.pop();
    taken++;
    for (int x : adj[t])
      if (--id[x] == 0) que.push(x);
  }
  if (taken < n) {
    printf("-1\n");
    return 0;
  }
  memset(memo, -1, sizeof memo);
  int h = 0;
  for (int i = 0; i < n; i++)
    for (char c = 'a'; c <= 'z'; c++) h = max(h, solve(i, c));
  printf("%d\n", h);
  return 0;
}
