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;
const int N = 3e5;
int n, m, k, flag[N + 10], now, dp[N + 10][30];
char s[N + 10];
vector<int> g[N + 10];
int ru[N + 10], a[N + 10];
queue<int> dl;
int main() {
  ios::sync_with_stdio(0), cin.tie(0);
  cin >> n >> m;
  cin >> (s + 1);
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    ru[y]++;
  }
  for (int i = 1; i <= n; i++) {
    a[i] = s[i] - 'a' + 1;
    if (ru[i] == 0) {
      dp[i][a[i]]++;
      dp[i][0] = 1;
    }
  }
  int num = 0;
  for (int i = 1; i <= n; i++)
    if (ru[i] == 0) {
      num++;
      dl.push(i);
      ru[i] = -1;
    }
  while (!dl.empty()) {
    int x = dl.front();
    dl.pop();
    for (int y : g[x]) {
      if (ru[y] <= 0) continue;
      if (dp[y][0] == 0) {
        for (int i = 1; i <= 26; i++) dp[y][i] = dp[x][i] + (a[y] == i);
      } else {
        for (int i = 1; i <= 26; i++)
          dp[y][i] = max(dp[y][i], dp[x][i] + (a[y] == i));
      }
      dp[y][0] = 1;
      ru[y]--;
      if (ru[y] == 0) {
        num++;
        dl.push(y);
      }
    }
  }
  if (num != n) {
    cout << -1 << endl;
    return 0;
  }
  int ma = 0;
  for (int i = 1; i <= n; i++)
    if (dp[i][0]) {
      for (int j = 1; j <= 26; j++) ma = max(ma, dp[i][j]);
    }
  cout << ma << endl;
  return 0;
}
