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> sortaret;
vector<int> G[300005];
int gr[300005];
int N, M;
string a;
queue<int> Q;
int pasi;
int rez;
int dp[300005];
int rundp(int c) {
  int rez = 0;
  memset(dp, 0, sizeof(dp));
  for (auto it : sortaret) {
    dp[it] = (a[it] == c + 'a');
    for (auto it2 : G[it]) {
      dp[it] = max(dp[it], dp[it2] + (a[it] == c + 'a'));
    }
    rez = max(rez, dp[it]);
  }
  return rez;
}
int main() {
  cin >> N >> M;
  cin >> a;
  a = " " + a;
  while (M--) {
    int x, y;
    cin >> x >> y;
    gr[y]++;
    G[x].push_back(y);
  }
  for (int i = 1; i <= N; i++) {
    if (!gr[i]) {
      Q.push(i);
    }
  }
  while (!Q.empty()) {
    int nod = Q.front();
    Q.pop();
    sortaret.push_back(nod);
    pasi++;
    for (auto it : G[nod]) {
      gr[it]--;
      if (!gr[it]) {
        Q.push(it);
      }
    }
  }
  if (pasi < N) {
    cout << -1;
    return 0;
  }
  reverse(sortaret.begin(), sortaret.end());
  for (int i = 0; i < 26; i++) {
    rez = max(rez, rundp(i));
  }
  cout << rez;
  return 0;
}
