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;
template <typename T>
void printvec(vector<T>& vec) {
  for (int i = 0; i < vec.size(); i++) {
    cout << vec[i] << " ";
  }
  cout << "\n";
}
int n, m;
string str;
vector<vector<int> > adj;
vector<bool> visited(300 * 1000 + 5, false);
vector<int> topo;
vector<set<int> > gr;
vector<vector<int> > ansarr(300 * 1000 + 5, vector<int>(26, 0));
void findans(int curr) {
  int i, j;
  set<int>::iterator it;
  for (it = gr[curr].begin(); it != gr[curr].end(); it++) {
    int x = *it;
    if (!visited[x]) {
      for (j = 0; j < 26; j++) {
        ansarr[curr][j] = max(ansarr[curr][j], ansarr[x][j]);
      }
    }
  }
  ansarr[curr][str[curr] - 'a']++;
}
int main() {
  std::ios::sync_with_stdio(false);
  int ttt = 1;
  while (ttt--) {
    int i, j;
    cin >> n >> m >> str;
    adj.resize(n);
    gr.resize(n);
    vector<int> indeg(n, 0);
    for (i = 0; i < m; i++) {
      int a, b;
      cin >> a >> b;
      a--;
      b--;
      gr[a].insert(b);
      if (a == b) {
        cout << "-1";
        return 0;
      }
      adj[a].push_back(b);
      indeg[b]++;
    }
    set<int> sett;
    for (i = 0; i < n; i++) {
      if (indeg[i] == 0) sett.insert(i);
    }
    while (!sett.empty()) {
      int curr = *sett.begin();
      sett.erase(curr);
      topo.push_back(curr);
      for (i = 0; i < adj[curr].size(); i++) {
        indeg[adj[curr][i]]--;
        if (indeg[adj[curr][i]] == 0) sett.insert(adj[curr][i]);
      }
    }
    if (topo.size() != n) {
      cout << "-1";
      return 0;
    }
    for (i = 0; i < n; i++) {
      visited[i] = false;
    }
    for (i = n - 1; i >= 0; i--) {
      if (!visited[topo[i]]) {
        findans(topo[i]);
      }
    }
    int ans = 0;
    for (i = 0; i < n; i++) {
      for (j = 0; j < 26; j++) {
        ans = max(ans, ansarr[i][j]);
      }
    }
    cout << ans;
  }
  return 0;
}
