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;
char letter[300005];
vector<int> graf[300005];
int deg[300005];
int dp[300005][26];
int main() {
  int n, m;
  cin >> n >> m;
  memset(deg, 0, sizeof deg);
  string s;
  cin >> s;
  for (int i = 0; i < s.size(); i++) letter[i + 1] = s[i];
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    graf[u].push_back(v);
    deg[v]++;
  }
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (deg[i] == 0) {
      dp[i][letter[i] - 'a'] = 1;
      q.push(i);
    }
  }
  int cnt = n;
  while (!q.empty()) {
    int front = q.front();
    q.pop();
    cnt--;
    for (auto child : graf[front]) {
      for (int i = 0; i < 26; i++)
        dp[child][i] =
            max(dp[child][i], (letter[child] - 'a' == i) + dp[front][i]);
      deg[child] -= 1;
      if (deg[child] == 0) q.push(child);
    }
  }
  if (cnt)
    cout << "-1";
  else {
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < 26; j++) {
        ans = max(ans, dp[i][j]);
      }
    }
    cout << ans;
  }
  return 0;
}
