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> graph[N];
int n, m, col[N], dp[N][26];
char str[N];
void cycle(int u) {
  if (col[u]) return;
  col[u] = 1;
  for (int v : graph[u]) {
    if (col[v] == 1) {
      printf("-1\n"), exit(0);
    }
    cycle(v);
  }
  col[u] = 2;
}
int dfs(int i, int ch) {
  int &ret = dp[i][ch];
  if (ret != -1) return ret;
  ret = 0;
  for (int v : graph[i]) {
    ret = max(ret, dfs(v, ch));
  }
  if (ch == str[i] - 'a') ret++;
  return ret;
}
int main() {
  scanf("%d %d", &n, &m);
  getchar();
  for (int i = 1; i <= n; i++) {
    scanf("%c", &str[i]);
  }
  int u, v;
  for (int i = 1; i <= m; i++) {
    scanf("%d %d", &u, &v);
    graph[u].push_back(v);
  }
  memset(col, 0, sizeof col);
  for (int i = 1; i <= n; i++) {
    if (!col[i]) cycle(i);
  }
  memset(dp, -1, sizeof dp);
  int result = -1;
  for (int i = 1; i <= n; i++) {
    for (int ch = 0; ch < 26; ch++) {
      result = max(result, dfs(i, ch));
    }
  }
  printf("%d\n", result);
}
