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;
map<int, int> mapa_1[1000000 + 100];
int n, a[1000000 + 100], sol, m;
string s;
set<pair<int, int>> se;
bool obidjen[1000000 + 100];
vector<int> v[1000000 + 100];
int mapa[1000000 + 100];
int res;
int bio[1000000 + 100];
int dp[300000 + 100][30];
queue<int> q;
void dfs(int x) {
  obidjen[x] = 1;
  for (auto y : v[x]) {
    if (obidjen[y]) continue;
    dfs(y);
  }
  bool o = false;
  for (auto y : v[x]) {
    o = true;
    for (int i = 'a'; i <= 'z'; i++) {
      if (i == s[x - 1])
        dp[x][i - 'a'] = max(dp[x][i - 'a'], 1 + dp[y][i - 'a']);
      else
        dp[x][i - 'a'] = max(dp[x][i - 'a'], dp[y][i - 'a']);
      res = max(res, dp[x][i - 'a']);
    }
  }
  if (!o) dp[x][s[x - 1] - 'a'] = 1;
}
void ciklus() {
  for (int i = 1; i <= n; i++)
    for (auto y : v[i]) bio[y]++;
  for (int i = 1; i <= n; i++)
    if (!bio[i]) q.push(i);
  int res = 0;
  while (!q.empty()) {
    int x = q.front();
    q.pop();
    res++;
    for (auto y : v[x]) {
      bio[y]--;
      if (bio[y] == 0) q.push(y);
    }
  }
  if (res < n) {
    cout << -1 << endl;
    exit(0);
  }
}
int main() {
  scanf("%d %d", &n, &m);
  cin >> s;
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    if (!mapa_1[x][y]) v[x].push_back(y), mapa_1[x][y] = 1;
  }
  ciklus();
  int maxx = 0;
  for (int i = 1; i <= n; i++) {
    if (!obidjen[i]) {
      se.clear();
      res = 0;
      se.insert({1, s[i - 1]});
      mapa[s[i - 1]] = 1;
      dfs(i);
      maxx = max(res, maxx);
    }
  }
  cout << maxx << endl;
  return 0;
}
