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;
long long int lim = 300000 + 10;
int mod = 1000000000 + 7;
int inf = 1000000000;
vector<vector<long long int>> path(lim);
vector<long long int> color(lim, 0);
bool pos = 0;
vector<bool> visited(lim, 0);
vector<long long int> ans;
string s;
void cycle(long long int v) {
  color[v] = 1;
  for (long long int u : path[v]) {
    if (color[u] == 1) {
      pos = 1;
    } else if (color[u] == 0) {
      cycle(u);
    }
  }
  color[v] = 2;
}
void dfs(long long int v) {
  visited[v] = 1;
  for (long long int u : path[v]) {
    if (!visited[u]) {
      dfs(u);
    }
  }
  ans.push_back(v);
}
void top_sort(int n) {
  for (int i = 1; i <= n; i++) {
    if (!visited[i]) {
      dfs(i);
    }
  }
  reverse(ans.begin(), ans.end());
}
long long int max(long long int u, long long int v) {
  if (u >= v) {
    return u;
  } else {
    return v;
  }
}
void countfn(long long int v, vector<bool> &counted, vector<long long int> &c,
             int i) {
  counted[v] = 1;
  if (s[v - 1] == 'a' + i) {
    c[v] = 1;
  } else {
    c[v] = 0;
  }
  long long int maxc = 0;
  for (long long int u : path[v]) {
    if (!counted[u]) {
      countfn(u, counted, c, i);
    }
    maxc = max(maxc, c[u]);
  }
  c[v] += maxc;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n, m;
  cin >> n >> m;
  cin >> s;
  for (int i = 0; i < m; i++) {
    int x, y;
    cin >> x >> y;
    path[x].push_back(y);
  }
  for (int i = 1; i <= n; i++) {
    if (color[i] != 2) {
      cycle(i);
    }
  }
  if (pos)
    cout << -1;
  else {
    long long int ans1 = 0;
    top_sort(n);
    for (int i = 0; i < 26; i++) {
      vector<long long int> c(n + 1);
      vector<bool> counted(n + 1);
      for (int j = 1; j <= n; j++) {
        if (path[j].size() == 0 && s[j - 1] == 'a' + i) {
          c[j] = 1;
          counted[j] = 1;
        }
      }
      for (int j = 0; j < n; j++) {
        if (!counted[ans[j]]) {
          countfn(ans[j], counted, c, i);
        }
      }
      long long int res = 0;
      for (int j = 1; j <= n; j++) {
        res = max(res, c[j]);
      }
      ans1 = max(ans1, res);
    }
    cout << ans1;
  }
}
