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;
int n, m, maxx;
bool CycleFound;
vector<int> v[300002];
map<long long, int> mp;
char lit[300002];
int color[300002];
bool DFSUtil(int nod) {
  color[nod] = 1;
  for (int i = 0; i < v[nod].size(); ++i) {
    if (color[v[nod][i]] == 1) return true;
    if (!color[v[nod][i]] && DFSUtil(v[nod][i])) return true;
  }
  color[nod] = 2;
  return false;
}
bool isCycle() {
  for (int i = 1; i <= n; ++i)
    if (!color[i] && DFSUtil(i)) return 1;
  return 0;
}
bool viz[300002];
bool viz2[300002];
int frq[300002][27];
int mi;
int fr[27];
void dfs(int nod) {
  viz[nod] = 1;
  for (int i = 0; i < v[nod].size(); ++i) {
    if (!viz[v[nod][i]]) dfs(v[nod][i]);
    for (int j = 1; j <= 26; ++j) {
      int nr = (lit[nod] == ('a' + j - 1));
      frq[nod][j] = max(frq[nod][j], frq[v[nod][i]][j] + nr);
    }
  }
  for (int i = 1; i <= 26; ++i) {
    int nr = (lit[nod] == ('a' + i - 1));
    frq[nod][i] = max(frq[nod][i], nr);
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n >> m;
  cin >> lit + 1;
  for (int i = 1; i <= m; ++i) {
    int a, b;
    cin >> a >> b;
    if (!mp[a * 1e6 + b]) {
      v[a].push_back(b);
      mp[a * 1e6 + b] = 1;
    }
  }
  if (isCycle())
    cout << -1;
  else {
    for (int i = 1; i <= n; ++i)
      if (!viz[i]) dfs(i);
    int maxn = 0;
    for (int j = 1; j <= n; ++j)
      for (int q = 1; q <= 26; ++q) maxn = max(maxn, frq[j][q]);
    cout << maxn;
  }
  return 0;
}
