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 maxn = 300005;
struct node {
  int v, next;
} eg[maxn * 2];
queue<int> que;
int head[maxn];
int in[maxn];
int dp[maxn][27];
int n, m;
int cnt;
char str[maxn];
void init() {
  memset(head, -1, sizeof(head));
  memset(in, 0, sizeof(in));
  cnt = 0;
}
void addeage(int u, int v) {
  eg[cnt].v = v;
  eg[cnt].next = head[u];
  head[u] = cnt++;
}
void topusort() {
  for (int i = 1; i <= n; i++) {
    if (in[i] == 0) {
      que.push(i);
      dp[i][str[i] - 'a']++;
    }
  }
  int num = 0;
  while (!que.empty()) {
    int u = que.front();
    que.pop();
    num++;
    for (int i = head[u]; i != -1; i = eg[i].next) {
      int v = eg[i].v;
      in[v]--;
      if (in[v] == 0) que.push(v);
      for (int j = 0; j < 26; j++) {
        if (str[v] - 'a' == j)
          dp[v][j] = max(dp[v][j], dp[u][j] + 1);
        else
          dp[v][j] = max(dp[v][j], dp[u][j]);
      }
    }
  }
  if (num < n) {
    printf("-1\n");
    return;
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 26; j++) {
      ans = max(ans, dp[i][j]);
    }
  }
  printf("%d\n", ans);
  return;
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", str + 1);
  init();
  int u, v;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    in[v]++;
    addeage(u, v);
  }
  topusort();
  return 0;
}
