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;
int n, m;
int namenod[300007];
vector<vector<int>> adj;
vector<int> topological;
int color[300007] = {0};
string s;
int dp[300007][29];
bool nocyc(int x) {
  bool b = true;
  for (auto nxt : adj[x]) {
    if (color[nxt] == 0)
      color[nxt] = 1, b &= nocyc(nxt);
    else if (color[nxt] == 1) {
      return 0;
    }
  }
  color[x] = 2;
  return b;
}
int dfs(int nod, int c) {
  if (adj[nod].size() == 0)
    return s[nod] - 'a' == c ? dp[nod][c] = 1 : dp[nod][c] = 0;
  if (dp[nod][c] != -1) return dp[nod][c];
  int greed = 0;
  for (auto nxt : adj[nod]) {
    greed = max(greed, dfs(nxt, c));
  }
  return s[nod] - 'a' == c ? dp[nod][c] = greed + 1 : dp[nod][c] = greed;
}
int main() {
  memset(dp, -1, sizeof dp);
  cin >> n >> m;
  adj.resize(n + 3);
  cin >> s;
  s = "0" + s;
  int x, y;
  for (int i = 0; i < m; ++i) {
    scanf("%d%d", &x, &y);
    adj[x].push_back(y);
  }
  for (int r = 1; r <= n; ++r) {
    if (!nocyc(r)) return puts("-1"), 0;
  }
  int ans = 0;
  for (int c = 0; c < 26; ++c) {
    for (int i = 1; i <= n; ++i) {
      ans = max(ans, dfs(i, c));
    }
  }
  cout << ans;
}
