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<int> v[300005], dp[300005];
string s;
int to[300005], n, m, CLK, RES;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n >> m >> s;
  s = '#' + s;
  for (int x, y; m--; v[x].push_back(y), ++to[y]) cin >> x >> y;
  queue<int> q;
  for (int i = 1; i <= n; dp[i++].resize(26, 0))
    if (!to[i]) q.push(i);
  for (; !q.empty(); ++CLK) {
    int u = q.front();
    ++dp[u][s[u] - 'a'];
    q.pop();
    RES = max(RES, *max_element(dp[u].begin(), dp[u].end()));
    for (auto i : v[u]) {
      for (int j = 0; j < 26; ++j) dp[i][j] = max(dp[i][j], dp[u][j]);
      if (!--to[i]) q.push(i);
    }
  }
  cout << (CLK != n ? -1 : RES);
}
