#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 5;
int ans;
string c;
bool cycle, in_stack[N];
vector<int> adj[N], vis[N];
void dfs(int node) {
  vis[node] = vector<int>(26, 0);
  in_stack[node] = 1;
  for (auto child : adj[node]) {
    if (in_stack[child]) cycle = 1;
    if (vis[child].empty()) dfs(child);
    for (int i = 0; i < 26; ++i)
      vis[node][i] = max(vis[node][i], vis[child][i]);
  }
  in_stack[node] = 0;
  ++vis[node][c[node - 1] - 'a'];
  for (int i = 0; i < 26; ++i) ans = max(ans, vis[node][i]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m >> c;
  while (m--) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
  }
  for (int i = 1; i <= n; ++i)
    if (vis[i].empty()) dfs(i);
  cout << (cycle ? -1 : ans);
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
const int N = 3e5 + 5;
int ans;
string c;
bool cycle, in_stack[N];
vector<int> adj[N], vis[N];
void dfs(int node) {
  vis[node] = vector<int>(26, 0);
  in_stack[node] = 1;
  for (auto child : adj[node]) {
    if (in_stack[child]) cycle = 1;
    if (vis[child].empty()) dfs(child);
    for (int i = 0; i < 26; ++i)
      vis[node][i] = max(vis[node][i], vis[child][i]);
  }
  in_stack[node] = 0;
  ++vis[node][c[node - 1] - 'a'];
  for (int i = 0; i < 26; ++i) ans = max(ans, vis[node][i]);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  int n, m;
  cin >> n >> m >> c;
  while (m--) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
  }
  for (int i = 1; i <= n; ++i)
    if (vis[i].empty()) dfs(i);
  cout << (cycle ? -1 : ans);
  return 0;
}

```