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>
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
using namespace std;
using ll = long long;
using lld = long double;
const int oo = 0x3f3f3f3f;
const ll MOD = 1000000007;
const int N = 3e5 + 10;
int onstack[N], dp[N][26];
bool vis[N];
vector<int> adj[N];
deque<int> dq;
bool cycle = 0;
void dfs(int src) {
  onstack[src] = 1;
  vis[src] = 1;
  for (int i : adj[src]) {
    if (!vis[i])
      dfs(i);
    else if (onstack[i])
      cycle = 1;
  }
  dq.push_front(src);
  onstack[src] = 0;
}
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int n, m;
  cin >> n >> m;
  string ar;
  cin >> ar;
  for (int i = 0; i < m; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    --v;
    adj[u].emplace_back(v);
  }
  for (int i = 0; i < n; i++)
    if (!vis[i]) dfs(i);
  if (cycle) {
    cout << "-1\n";
    return 0;
  }
  auto relax = [](int& a, int b) { a = max(a, b); };
  for (int i : dq) {
    dp[i][ar[i] - 'a']++;
    for (int j : adj[i]) {
      for (int c = 0; c < 26; c++) relax(dp[j][c], dp[i][c]);
    }
  }
  int ans = 0;
  for (int i = 0; i < n; i++) {
    for (int j = 0; j < 26; j++) ans = max(ans, dp[i][j]);
  }
  cout << ans << "\n";
  return 0;
}
