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 N = 300005;
int dp[N][26] = {0};
vector<int> g[N];
char cl[N] = {0};
int n, m;
string line;
int answer = 0;
int ctoi(char c) { return c - 'a'; }
bool dfs(int v) {
  int cIndex = ctoi(line[v]);
  cl[v] = 1;
  dp[v][cIndex] = 1;
  for (int to : g[v]) {
    if (cl[to] == 0) {
      if (dfs(to)) return true;
    } else if (cl[to] == 1) {
      return true;
    }
    for (int i = 0; i < 26; ++i) {
      int add = i == cIndex;
      dp[v][i] = max(dp[v][i], dp[to][i] + add);
      answer = max(answer, dp[v][i]);
    }
  }
  answer = max(answer, dp[v][cIndex]);
  cl[v] = 2;
  return false;
}
int main(int argc, const char* argv[]) {
  ios::sync_with_stdio(false);
  cin >> n >> m;
  cin.get();
  getline(cin, line);
  for (int i = 0; i < m; ++i) {
    int a, b;
    cin >> a >> b;
    g[a - 1].push_back(b - 1);
  }
  for (int i = 0; i < n; ++i) {
    if (!cl[i]) {
      if (dfs(i)) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  cout << answer << endl;
  return 0;
}
