#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  s = '#' + s;
  int dp[n + 1][26], d[n + 1];
  queue<int> q;
  memset(dp, 0, sizeof(dp));
  vector<int> v[n + 1];
  memset(d, 0, sizeof(d));
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    d[b]++;
  }
  for (int i = 1; i <= n; i++) {
    if (!d[i]) {
      q.push(i);
      dp[i][s[i] - 'a'] = 1;
    }
  }
  int r = n;
  while (!q.empty()) {
    r--;
    int t = q.front();
    q.pop();
    for (auto i : v[t]) {
      for (int j = 0; j < 26; j++) {
        dp[i][j] = max(dp[i][j], dp[t][j] + (s[i] - 'a' == j));
      }
      d[i]--;
      if (!d[i]) {
        q.push(i);
      }
    }
  }
  if (r) {
    cout << -1 << "\n";
  } else {
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < 26; j++) ans = max(ans, dp[i][j]);
    }
    cout << ans;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m;
  string s;
  cin >> s;
  s = '#' + s;
  int dp[n + 1][26], d[n + 1];
  queue<int> q;
  memset(dp, 0, sizeof(dp));
  vector<int> v[n + 1];
  memset(d, 0, sizeof(d));
  for (int i = 0; i < m; i++) {
    int a, b;
    cin >> a >> b;
    v[a].push_back(b);
    d[b]++;
  }
  for (int i = 1; i <= n; i++) {
    if (!d[i]) {
      q.push(i);
      dp[i][s[i] - 'a'] = 1;
    }
  }
  int r = n;
  while (!q.empty()) {
    r--;
    int t = q.front();
    q.pop();
    for (auto i : v[t]) {
      for (int j = 0; j < 26; j++) {
        dp[i][j] = max(dp[i][j], dp[t][j] + (s[i] - 'a' == j));
      }
      d[i]--;
      if (!d[i]) {
        q.push(i);
      }
    }
  }
  if (r) {
    cout << -1 << "\n";
  } else {
    int ans = 0;
    for (int i = 1; i <= n; i++) {
      for (int j = 0; j < 26; j++) ans = max(ans, dp[i][j]);
    }
    cout << ans;
  }
  return 0;
}

```