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;
vector<long long int> adj[300005];
bool visit[300005];
bool cyc[300005];
long long int dp[300005][26], fl = 0;
string s;
void dfs(long long int x) {
  visit[x] = true;
  cyc[x] = true;
  for (auto it : adj[x]) {
    if (cyc[it]) fl = 1;
    if (!visit[it]) dfs(it);
  }
  for (auto it : adj[x]) {
    for (long long int i = 0; i < 26; i++) {
      dp[x][i] = max(dp[x][i], dp[it][i]);
    }
  }
  long long int p = s[x - 1] - 'a';
  dp[x][p]++;
  cyc[x] = false;
}
void solve() {
  long long int n, m;
  cin >> n >> m;
  cin >> s;
  for (long long int i = 0; i < m; i++) {
    long long int x, y;
    cin >> x >> y;
    adj[x].push_back(y);
  }
  for (long long int i = 0; i < 300004; i++) {
    visit[i] = false;
    cyc[i] = false;
    for (long long int j = 0; j < 26; j++) dp[i][j] = 0;
  }
  for (long long int i = 1; i <= n; i++) {
    if (!visit[i]) {
      dfs(i);
    }
  }
  if (fl) {
    cout << -1 << endl;
    return;
  }
  long long int ans = 0;
  for (long long int i = 1; i < n + 1; i++) {
    for (long long int j = 0; j < 26; j++) ans = max(ans, dp[i][j]);
  }
  cout << ans << endl;
}
signed main() {
  ios_base::sync_with_stdio(false);
  long long int t = 1;
  while (t--) solve();
  return 0;
}
