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 long long int N = 3e5 + 5;
const long long int INF = 1e18;
const long long int mod = 1e9 + 7;
string s;
long long int dp[N][27];
long long int n, m, u, v;
vector<long long int> g[N];
long long int vis[N], maxi = 0, col[N], visx[N];
void DFS(long long int cur, long long int par) {
  vis[cur] = 1;
  dp[cur][int(s[cur - 1]) - 97]++;
  for (long long int j : g[cur]) {
    if (!vis[j]) {
      DFS(j, cur);
    }
    for (long long int i = 0; i < 26; i++) {
      if (i == int(s[cur - 1]) - 97)
        dp[cur][i] = max(dp[cur][i], dp[j][i] + 1);
      else
        dp[cur][i] = max(dp[cur][i], dp[j][i]);
    }
  }
  for (long long int i = 0; i < 26; i++) {
    maxi = max(maxi, dp[cur][i]);
  }
}
bool DFSX(long long int cur, long long int par) {
  col[cur] = -1;
  for (long long int i : g[cur]) {
    if (col[i] == -1) {
      return true;
    }
    if (col[i] == 0 && DFSX(i, cur)) {
      return true;
    }
  }
  col[cur] = 1;
  return false;
}
int main() {
  cin >> n >> m;
  cin >> s;
  while (m--) {
    cin >> u >> v;
    g[u].push_back(v);
  }
  for (long long int i = 1; i <= n; i++) {
    if (!visx[i]) {
      if (DFSX(i, -1)) {
        cout << -1 << endl;
        exit(0);
      }
    }
  }
  for (long long int i = 1; i <= n; i++) {
    if (!vis[i]) {
      DFS(i, 0);
    }
  }
  cout << maxi << endl;
}
