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 V(300011), E(300011), S(27);
int h[V], n[E], e[E], vt, et, a[V], id[V], mx[S][V], q[V], ql, qr;
void ins(int f, int t) {
  n[++et] = h[f];
  h[f] = et;
  e[et] = t;
  ++id[t];
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int m, i, j, k, tot, ans;
  char ch;
  cin >> vt >> m;
  for (i = 1; i <= vt; ++i) {
    cin >> ch;
    a[i] = ch - 'a';
  }
  while (m--) {
    cin >> i >> j;
    ins(i, j);
  }
  ql = qr = 0;
  for (i = 1; i <= vt; ++i) {
    mx[a[i]][i] = 1;
    if (!id[i]) q[qr++] = i;
  }
  ans = tot = 0;
  while (ql < qr) {
    ++tot;
    j = q[ql++];
    ans = max(ans, mx[a[j]][j]);
    for (i = h[j]; i; i = n[i]) {
      for (k = 0; k < 26; ++k)
        mx[k][e[i]] = max(mx[k][e[i]], mx[k][j] + (k == a[e[i]]));
      if (!--id[e[i]]) q[qr++] = e[i];
    }
  }
  if (tot < vt) ans = -1;
  cout << ans << endl;
  return 0;
}
