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, ans;
int p[300005], f[300005][26];
vector<long long> adj[300005];
string s;
queue<int> q;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> s;
  while (m--) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    p[v]++;
  }
  for (long long i = 1; i < n + 1; ++i)
    if (!p[i]) q.push(i);
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    n--;
    f[x][s[x - 1] - 'a']++;
    for (long long i = 0; i < 26; ++i)
      if (ans < f[x][i]) ans = f[x][i];
    for (long long i = 0; i < (long long)adj[x].size(); ++i) {
      int y = adj[x][i];
      for (long long j = 0; j < 26; ++j) f[y][j] = max(f[y][j], f[x][j]);
      p[y]--;
      if (!p[y]) q.push(y);
    }
  }
  if (n > 0) ans = -1;
  cout << ans;
  return 0;
}
