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<bool> ancestor(300005, false);
vector<bool> visited(300005, false);
int countmap[300005][26];
string s;
int ans = 0;
void dfs(int ind, vector<list<int>> &adj, bool &has_cycle) {
  ancestor[ind] = true;
  visited[ind] = true;
  for (auto it = adj[ind].begin(); it != adj[ind].end(); it++) {
    if (!visited[*it])
      dfs(*it, adj, has_cycle);
    else {
      if (ancestor[*it]) has_cycle = true;
    }
    for (int i = 0; i < 26; i++)
      countmap[ind][i] = max(countmap[ind][i], countmap[*it][i]);
  }
  countmap[ind][s[ind] - 97]++;
  ancestor[ind] = false;
}
int main() {
  int n, m;
  scanf("%d%d", &n, &m);
  cin >> s;
  vector<list<int>> adj(n);
  for (int i = 0; i < m; i++) {
    int a, b;
    scanf("%d%d", &a, &b);
    adj[a - 1].push_back(b - 1);
  }
  for (int i = 0; i < n; i++) {
    bool has_cycle = false;
    if (!visited[i]) dfs(i, adj, has_cycle);
    if (has_cycle) {
      ans = -1;
      break;
    }
    for (int j = 0; j < 26; j++) ans = max(ans, countmap[i][j]);
  }
  printf("%d\n", ans);
  return 0;
}
