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 N = 300000 + 10;
const int M = 30;
int n, m, d[N][M], vis[N], tip[N];
string s;
vector<int> G[N];
bool dfs(int u) {
  vis[u] = -1;
  for (int v : G[u]) {
    if (vis[v] == -1) return false;
    if (!vis[v] && !dfs(v)) return false;
  }
  vis[u] = 1;
  return true;
}
int dp(int u, int alpha) {
  int& ans = d[u][alpha];
  if (ans != -1) return ans;
  ans = 0;
  for (int v : G[u]) ans = max(ans, dp(v, alpha));
  ans += (tip[u] == alpha);
  return ans;
}
int main() {
  memset(d, -1, sizeof(d));
  std::ios_base::sync_with_stdio(false);
  cin.tie(0);
  cin >> n >> m;
  cin >> s;
  for (int i = 0; i < n; i++) tip[i] = s[i] - 'a';
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    G[x - 1].push_back(y - 1);
  }
  for (int i = 0; i < n; i++)
    if (!vis[i] && !dfs(i)) {
      cout << -1 << '\n';
      return 0;
    }
  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 << '\n';
  return 0;
}
