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 + 5;
int n, m, k, l, res;
vector<int> arr[N];
string s;
int dp[N][26];
int visited[N];
void dfs(int cur) {
  if (visited[cur]) return;
  dp[cur][s[cur - 1] - 'a']++;
  visited[cur] = 1;
  for (int i = 0; i < arr[cur].size(); i++) {
    if (visited[arr[cur][i]] == 1)
      k = -1;
    else
      dfs(arr[cur][i]);
  }
  for (int j = 0; j < 26; j++) {
    int curRes = 0;
    for (int i = 0; i < arr[cur].size(); i++) {
      if (dp[arr[cur][i]][j] > curRes) {
        curRes = dp[arr[cur][i]][j];
      }
    }
    dp[cur][j] += curRes;
    if (res < dp[cur][j]) {
      res = dp[cur][j];
    }
  }
  visited[cur] = 2;
}
int main() {
  cin >> n >> m;
  cin >> s;
  for (int i = 0; i < m; ++i) {
    cin >> k >> l;
    arr[k].push_back(l);
  }
  for (int i = 1; i <= n; i++) {
    dfs(i);
  }
  if (k == -1) res = -1;
  cout << res << endl;
  return 0;
}
