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> v[300005];
string h;
stack<long long int> s;
set<long long int> st;
long long int c[300005][26], a[300005], x, y, n, m, res = 0;
bool vis[300005];
void dfs(int pos) {
  vis[pos] = true;
  st.insert(pos);
  for (auto u : v[pos])
    if (st.find(u) != st.end()) {
      cout << -1 << "\n";
      exit(0);
    } else if (!vis[u])
      dfs(u);
  st.erase(pos);
  s.push(pos);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m >> h;
  for (int i = 0; i < h.size(); i++) a[i + 1] = h[i] - 'a';
  for (int i = 0; i < m; i++) cin >> x >> y, v[x].push_back(y);
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs(i);
  while (!s.empty()) {
    x = s.top();
    s.pop();
    c[x][a[x]]++;
    for (auto u : v[x])
      for (int i = 0; i < 26; i++) c[u][i] = max(c[u][i], c[x][i]);
    for (int i = 0; i < 26; i++) res = max(res, c[x][i]);
  }
  cout << res << "\n";
  return 0;
}
