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 MIN = 2e3 + 2;
const int MXN = 3e5 + 3;
const int INF = 1e9 + 7;
const int base = 1e9;
const long long LINF = 1e18 + 15;
const double EPS = 1e-9;
int n, m, mx;
char ch[MXN];
vector<int> g[MXN];
int u, v;
vector<int> g1[MXN];
int used[MXN];
bool used2[MXN][27];
int sum[MXN][27];
inline bool dfs(int x) {
  used[x] = 1;
  for (int i = 0; i < g[x].size(); ++i) {
    int to = g[x][i];
    if (!used[to]) {
      if (dfs(to)) return true;
    } else if (used[to] == 1) {
      return true;
    }
  }
  used[x] = 2;
  return false;
}
inline int dfs2(int x, char c) {
  if (used2[x][c - 'a'] == 1) return sum[x][c - 'a'];
  int ans = 0;
  used2[x][c - 'a'] = 1;
  for (int i = 0; i < g[x].size(); ++i) {
    int to = g[x][i];
    if (used2[to][c - 'a']) {
      ans = max(ans, sum[to][c - 'a']);
    } else {
      ans = max(ans, dfs2(to, c));
    }
  }
  if (ch[x] == c) {
    ans++;
  }
  sum[x][c - 'a'] = ans;
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; ++i) {
    cin >> ch[i];
  }
  for (int i = 1; i <= m; ++i) {
    cin >> u >> v;
    g[u].push_back(v);
    g1[v].push_back(u);
  }
  for (int i = 1; i <= n; ++i) {
    if (!used[i]) {
      if (dfs(i)) {
        cout << -1;
        return 0;
      }
    }
  }
  for (int i = 1; i <= n; ++i) {
    for (char j = 'a'; j <= 'z'; ++j) {
      if (!used2[i][j - 'a']) {
        mx = max(mx, dfs2(i, j));
      }
    }
  }
  cout << mx;
  return 0;
}
