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 double PI =
    3.141592653589793238462643383279502884197169399375105820974944;
int power(int a, int b, int m = 1000000007) {
  if (b == 0) return 1;
  if (b == 1) return a;
  int x = power(a, b / 2, m) % m;
  x = (x * x) % m;
  if (b % 2) return (x * a) % m;
  return x;
}
int n, m;
string s;
vector<int> adj[300005];
int vis[300005];
int deg[300005];
int dp[300005][30];
vector<int> v;
bool check() {
  queue<int> q;
  for (int i = 0; i < n; i++)
    if (!deg[i]) q.push(i);
  int cnt = q.size();
  while (!q.empty()) {
    int u = q.front();
    q.pop();
    v.push_back(u);
    for (auto v : adj[u]) {
      deg[v]--;
      if (!deg[v]) {
        cnt++;
        q.push(v);
      }
    }
  }
  return cnt != n;
}
void solve(int u) {
  vis[u] = 1;
  for (int i = 0; i < 26; i++) dp[u][i] = 0;
  for (auto v : adj[u]) {
    if (!vis[v]) solve(v);
    for (int i = 0; i < 26; i++) dp[u][i] = max(dp[u][i], dp[v][i]);
  }
  dp[u][s[u] - 'a']++;
}
signed main() {
  cin >> n >> m >> s;
  for (int i = 0; i < m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    u--;
    v--;
    deg[v]++;
    adj[u].push_back(v);
  }
  if (check()) {
    cout << -1;
    return 0;
  }
  for (int i = 0; i < n + 5; i++) vis[i] = 0;
  int ans = 0;
  for (auto i : v) {
    if (!vis[i]) solve(i);
    for (int j = 0; j < 26; j++) ans = max(ans, dp[i][j]);
  }
  cout << ans << endl;
}
