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;
int n, m, used[300005], dp[300005][30];
string s;
vector<int> g[300005];
void dfs(int v) {
  used[v] = 1;
  dp[v][s[v] - 'a'] = 1;
  int f[30];
  for (int i = 0; i < 26; i++) f[i] = 0;
  for (auto to : g[v]) {
    if (used[to] == 1) {
      cout << -1;
      exit(0);
    }
    if (used[to] == 0) {
      dfs(to);
    }
    for (int i = 0; i < 26; i++) f[i] = max(f[i], dp[to][i]);
  }
  for (int i = 0; i < 26; i++) dp[v][i] += f[i];
  used[v] = 2;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> s;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
  }
  for (int i = 0; i < n; i++)
    if (used[i] == 0) dfs(i);
  int ans = 0;
  for (int i = 0; i < n; i++)
    for (int j = 0; j < 26; j++) ans = max(ans, dp[i][j]);
  cout << ans;
}
