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;
vector<int> g[300005];
int cnt[300005];
int vis[300005];
int ans[300001][26];
int res = 0;
int main() {
  queue<int> q;
  int n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  int x, y;
  while (m--) {
    scanf("%d%d", &x, &y);
    x--, y--;
    g[x].push_back(y);
    cnt[y]++;
  }
  for (int i = 0; i < s.size(); i++) {
    ans[i][s[i] - 'a']++;
    if (!cnt[i]) q.push(i);
  }
  while (!q.empty()) {
    int i = q.front();
    q.pop();
    for (auto v : g[i]) {
      cnt[v]--;
      for (int j = 0; j < 26; j++) {
        ans[v][j] = max(ans[v][j], ans[i][j] + (s[v] == j + 'a'));
        res = max(res, ans[v][j]);
      }
      if (!cnt[v]) q.push(v);
    }
  }
  if (*max_element(cnt, cnt + s.size())) return cout << -1, 0;
  cout << res;
  return 0;
}
