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 N = (int)3e5 + 11;
int n, m, dp[N][30], cnt[N];
vector<int> g[N], q[N];
int u[N];
bool cyc;
string s = "#";
void dfs(int v) {
  u[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (u[to] == 1 or cyc) {
      cyc = 1;
      return;
    }
    if (u[to] == 0) dfs(to);
  }
  u[v] = 2;
}
int main() {
  ;
  ios_base::sync_with_stdio(0);
  cin.tie();
  cout.tie();
  ;
  cin >> n >> m;
  string q1;
  cin >> q1;
  s += q1;
  for (int i = 1; i <= m; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    q[y].push_back(x);
    if (x == y) cyc = 1;
    cnt[y]++;
  }
  bool op = 1;
  for (int i = 1; i <= n; i++) {
    if (u[i] == 0 and q[i].empty()) dfs(i), op = 0;
    if (cyc) {
      cout << "-1\n";
      return 0;
    }
  }
  if (op) {
    cout << "-1";
    return 0;
  }
  memset(u, 0, sizeof(u));
  queue<int> w;
  for (int i = 1; i <= n; i++)
    if (q[i].empty()) {
      u[i] = 1;
      w.push(i);
    }
  while (!w.empty()) {
    int v = w.front();
    bool ok = 1;
    for (int i = 0; i < q[v].size(); i++)
      for (int j = 0; j < 26; j++) {
        int to = q[v][i];
        dp[v][j] = max(dp[v][j], dp[to][j]);
      }
    dp[v][s[v] - 'a']++;
    for (int i = 0; i < g[v].size(); i++) {
      int to = g[v][i];
      if (!u[to]) {
        cnt[to]--;
        if (cnt[to]) continue;
        u[to] = 1;
        w.push(to);
      }
    }
    w.pop();
  }
  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;
}
