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 = 300005;
vector<int> G[N];
char col[N];
int n, m;
int dp[N];
int idx[N];
int ind[N];
int total;
int dfs(int u) {
  idx[u] = ++total;
  for (int i = 0; i < G[u].size(); ++i) {
    int v = G[u][i];
    if (idx[v] != 0) {
      return 1;
    }
    dfs(v);
  }
  return 0;
}
char ch;
int gao(int u) {
  if (dp[u] != -1) return dp[u];
  int ans = 0;
  for (int i = 0; i < G[u].size(); ++i) {
    int v = G[u][i];
    ans = max(ans, gao(v));
  }
  return dp[u] = (ch == col[u]) + ans;
}
void work() {
  scanf("%d%d", &n, &m);
  scanf("%s", col + 1);
  int self_circle = 0;
  for (int i = 0; i < m; ++i) {
    int a, b;
    scanf("%d%d", &a, &b);
    if (a == b) {
      self_circle = 1;
    }
    G[a].push_back(b);
    ind[b]++;
  }
  if (self_circle == 1) {
    puts("-1");
    return;
  }
  {
    queue<int> que;
    for (int i = 1; i <= n; ++i) {
      if (ind[i] == 0) que.push(i);
    }
    while (!que.empty()) {
      int u = que.front();
      que.pop();
      for (int i = 0; i < G[u].size(); ++i) {
        int v = G[u][i];
        if (--ind[v] == 0) {
          que.push(v);
        }
      }
    }
    for (int i = 1; i <= n; ++i) {
      if (ind[i] != 0) {
        puts("-1");
        return;
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < 26; ++i) {
    ch = 'a' + i;
    memset(dp, -1, sizeof(dp));
    for (int u = 1; u <= n; ++u) {
      ans = max(ans, gao(u));
    }
  }
  printf("%d\n", ans);
  return;
}
int main() {
  work();
  return 0;
}
