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 INF = 1e9 + 10, MN = 3e5 + 10;
short mk[MN];
string s;
vector<int> adj[MN], topol;
int dp[27][MN];
int n, m;
void fail() {
  cout << -1 << '\n';
  exit(0);
}
void dfs(int s) {
  mk[s] = 2;
  for (auto v : adj[s]) {
    if (!mk[v])
      dfs(v);
    else if (mk[v] == 2)
      fail();
  }
  mk[s] = 1;
  topol.push_back(s);
}
int main() {
  ios_base ::sync_with_stdio(false), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  cin >> s;
  for (int i = 0; i < m; ++i) {
    int a, b;
    ;
    cin >> a >> b;
    --a, --b;
    if (a == b) fail();
    adj[a].push_back(b);
  }
  for (int i = 0; i < n; ++i)
    if (!mk[i]) dfs(i);
  for (auto v : topol) {
    dp[s[v] - 'a'][v] = 1;
    for (int i = 0; i < 26; ++i) {
      for (auto u : adj[v])
        dp[i][v] = max(dp[i][v], (i == s[v] - 'a') + dp[i][u]);
    }
  }
  int ans = 0;
  for (int i = 0; i < n; ++i)
    for (int j = 0; j < 26; ++j) ans = max(ans, dp[j][i]);
  cout << ans << '\n';
  return 0;
}
