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 mod = 1e9 + 7;
set<pair<int, int> > st;
int n, m;
string second;
vector<int> adj[300001];
int fr[30];
bool vis[300001];
bool e[300001];
int dp[300001];
void dfs(int node) {
  if (!vis[node]) {
    e[node] = 1;
    vis[node] = 1;
    for (int i = 0; i < adj[node].size(); i++) {
      if (!e[adj[node][i]]) {
        dfs(adj[node][i]);
      } else {
        puts("-1");
        exit(0);
      }
    }
    e[node] = 0;
  }
  return;
}
int dfs2(int node, int x) {
  int &ret = dp[node];
  if (~ret) return ret;
  ret = 0;
  for (int i = 0; i < adj[node].size(); i++) {
    ret = max(ret, dfs2(adj[node][i], x));
  }
  return ret = ret + (second[node - 1] == x + 'a');
}
signed main() {
  int n, m;
  cin >> n >> m;
  cin >> second;
  int u, v;
  for (int i = 0; i < m; i++) {
    scanf("%d%d", &u, &v);
    st.insert({u, v});
  }
  vector<pair<int, int> > in(st.begin(), st.end());
  for (int i = 0; i < in.size(); i++) {
    adj[in[i].first].push_back(in[i].second);
  }
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    if (!vis[i]) {
      dfs(i);
    }
  }
  for (int i = 0; i < 26; i++) {
    memset(dp, -1, sizeof dp);
    for (int j = 1; j <= n; j++) {
      ans = max(ans, dfs2(j, i));
    }
  }
  cout << ans << endl;
  return 0;
}
