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;
char s[300005];
struct node {
  int to, next;
} edge[300005 * 2];
int head[300005], sumedge, ind[300005];
bool vis[300005];
int que[300005], ql, qr, dp[300005];
void add(int u, int v) {
  edge[++sumedge].to = v;
  edge[sumedge].next = head[u];
  head[u] = sumedge;
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", s + 1);
  int u, v;
  for (int i = 1; i <= m; ++i) {
    scanf("%d%d", &u, &v);
    add(u, v);
    ind[v]++;
  }
  qr = 0;
  ql = 1;
  for (int i = 1; i <= n; ++i) {
    if (ind[i] == 0) {
      que[++qr] = i;
      vis[i] = 1;
    }
  }
  while (ql <= qr) {
    int now = que[ql++];
    for (int i = head[now]; i; i = edge[i].next) {
      int v = edge[i].to;
      --ind[v];
      if (ind[v] == 0 && vis[v] == 0) {
        que[++qr] = v;
        vis[v] = 1;
      }
    }
  }
  if (qr < n) {
    puts("-1");
  } else {
    int ans = 0;
    for (int i = 0; i <= 25; ++i) {
      memset(dp, 0, sizeof(dp));
      for (int j = 1; j <= n; ++j) {
        int now = que[j];
        if (s[now] == 'a' + i) dp[now]++;
        for (int h = head[now]; h; h = edge[h].next) {
          int v = edge[h].to;
          dp[v] = max(dp[v], dp[now]);
        }
        ans = max(ans, dp[now]);
      }
    }
    printf("%d\n", ans);
  }
  return 0;
}
