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;
const long long mod = 1e9 + 7;
const long long INF = 1e9;
const double eps = 1e-9;
int n, m;
int degree[maxn];
char s[maxn];
int dp[maxn];
vector<int> vec[maxn];
vector<int> mp;
bool check() {
  mp.clear();
  queue<int> q;
  for (int i = 1; i <= n; i++) {
    if (degree[i] == 0) q.push(i);
  }
  while (q.size()) {
    int u = q.front();
    q.pop();
    mp.push_back(u);
    for (int i = 0; i < vec[u].size(); i++) {
      degree[vec[u][i]]--;
      if (degree[vec[u][i]] == 0) q.push(vec[u][i]);
    }
  }
  return mp.size() == n;
}
int solve(int x) {
  memset((dp), (0), sizeof(dp));
  char c = x + 'a';
  int cnt = 0;
  for (int i = mp.size() - 1; i >= 0; i--) {
    int u = mp[i];
    for (int j = 0; j < vec[u].size(); j++) {
      dp[u] = max(dp[u], dp[vec[u][j]]);
    }
    if (s[u - 1] == c) dp[u]++;
    cnt = max(cnt, dp[u]);
  }
  return cnt;
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    for (int i = 1; i <= n; i++) vec[i].clear();
    scanf("%s", s);
    memset((degree), (0), sizeof(degree));
    for (int i = 0; i < m; i++) {
      int x, y;
      scanf("%d%d", &x, &y);
      vec[x].push_back(y);
      degree[y]++;
    }
    if (!check()) {
      puts("-1");
      continue;
    }
    int ans = 0;
    for (int i = 0; i < 26; i++) {
      ans = max(ans, solve(i));
    }
    printf("%d\n", ans);
  }
}
