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;
pair<int, int> yol[(int)(3e5 + 5)];
int n, m, vis[(int)(3e5 + 5)], ti, dp[(int)(3e5 + 5)][30], ans;
char s[(int)(3e5 + 5)];
stack<int> st;
vector<int> v, ed[(int)(3e5 + 5)];
void dfs(int cur) {
  if (vis[cur]) return;
  vis[cur] = 1;
  for (auto i : ed[cur]) dfs(i);
  st.push(cur);
}
int main() {
  scanf("%d %d %s", &n, &m, s + 1);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d %d", &u, &v);
    ed[u].push_back(v);
    yol[i] = make_pair(u, v);
  }
  for (int i = 1; i <= n; i++) dfs(i);
  while ((int)st.size()) {
    vis[st.top()] = ++ti;
    v.push_back(st.top());
    st.pop();
  }
  for (int i = 1; i <= m; i++) {
    if (vis[yol[i].first] >= vis[yol[i].second]) return printf("-1"), 0;
  }
  for (int i = (int)v.size() - 1; i >= 0; i--) {
    int cur = v[i];
    for (auto x : ed[cur]) {
      for (int j = 0; j <= 26; j++) {
        dp[cur][j] = max(dp[cur][j], dp[x][j]);
      }
    }
    dp[cur][s[cur] - 'a']++;
    for (int j = 0; j <= 26; j++) {
      ans = max(dp[cur][j], ans);
    }
  }
  printf("%d\n", ans);
  return 0;
}
