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 = 3e5 + 10;
vector<int> v[maxn];
char s[maxn];
int n, m, dp[maxn], vis[maxn], flag;
void dfs1(int x) {
  if (vis[x] == 1) {
    flag = 1;
    return;
  }
  if (vis[x]) return;
  vis[x] = 1;
  for (auto i : v[x]) {
    dfs1(i);
  }
  vis[x] = 2;
}
void dfs2(int x, int y) {
  if (vis[x]) return;
  vis[x] = 1;
  int ans = 0;
  for (auto i : v[x]) {
    dfs2(i, y);
    ans = max(ans, dp[i]);
  }
  dp[x] = ans;
  if (s[x - 1] - 'a' == y) dp[x]++;
}
int main() {
  scanf("%d%d%s", &n, &m, s);
  for (int i = 0; i < m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) dfs1(i);
    if (flag) {
      puts("-1");
      return 0;
    }
  }
  int ans = 0;
  for (int j = 0; j < 26; j++) {
    for (int i = 1; i <= n; i++) dp[i] = vis[i] = 0;
    for (int i = 1; i <= n; i++) {
      if (!vis[i]) dfs2(i, j);
    }
    for (int i = 1; i <= n; i++) ans = max(ans, dp[i]);
  }
  printf("%d", ans);
  return 0;
}
