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 inf = (int)1e9;
const long long int INF = (long long int)2e18;
const int N = 100005;
const int mod = 1000000007;
const double eps = 1e-8;
const double pi = acos(-1.0);
int n, m;
vector<int> graph[3 * N];
string str;
int color[3 * N];
int res[3 * N][30];
int val[3 * N];
void iscycle(int s) {
  if (color[s] == 2) return;
  if (color[s] == 1) {
    cout << -1 << endl;
    exit(0);
  }
  color[s] = 1;
  for (auto x : graph[s]) iscycle(x);
  color[s] = 2;
}
void dfs(int s) {
  res[s][val[s]] = 1;
  color[s] = 1;
  for (auto x : graph[s]) {
    if (color[x] == 0) dfs(x);
    for (int i = 1; i <= 26; i++) {
      res[s][i] = max(res[s][i], res[x][i] + (i == (val[s])));
    }
  }
}
int main(int argc, char const *argv[]) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m;
  cin >> str;
  for (int i = 0; i < str.size(); i++) val[i + 1] = str[i] - 'a' + 1;
  for (int i = 1; i <= m; i++) {
    int u, v;
    cin >> u >> v;
    graph[u].push_back(v);
  }
  memset(color, 0, sizeof(color));
  for (int i = 1; i <= n; i++)
    if (color[i] == 0) iscycle(i);
  memset(color, 0, sizeof(color));
  for (int i = 1; i <= n; i++)
    if (color[i] == 0) dfs(i);
  int mx = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= 26; j++) {
      mx = max(mx, res[i][j]);
    }
  }
  cout << mx << endl;
  return 0;
}
