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>
#pragma comment(linker, "/STACK:210886422")
using namespace std;
const int MOD = 1000000007;
const int INF = 1000000001;
const long double EPS = 1e-13;
const int HASH_POW = 29;
const long double PI = acos(-1.0);
mt19937_64 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
long long nextLong(long long l, long long r) {
  uniform_int_distribution<long long> foo(l, r);
  return foo(rnd);
}
double nextDouble(double l, double r) {
  uniform_real_distribution<double> foo(l, r);
  return foo(rnd);
}
double workTime() { return double(clock()) / CLOCKS_PER_SEC; }
void myReturn(int code = 0) { exit(code); }
int n, m;
string s;
vector<int> G[300010];
int used[300010];
vector<int> ord;
bool haveCycle(int v) {
  used[v] = 1;
  for (int i = 0; i < G[v].size(); ++i) {
    int to = G[v][i];
    if (used[to] == 0) {
      if (haveCycle(to)) {
        return true;
      }
    } else if (used[to] == 1) {
      return true;
    }
  }
  used[v] = 2;
  ord.push_back(v);
  return false;
}
int dp[300010][26];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> n >> m >> s;
  while (m--) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
  }
  for (int i = 1; i <= n; ++i) {
    if (!used[i]) {
      if (haveCycle(i)) {
        cout << -1 << endl;
        myReturn();
      }
    }
  }
  for (int i = 0; i < n; ++i) {
    int v = ord[i];
    int cnt[26];
    for (int j = 0; j < 26; ++j) {
      dp[v][j] = 0;
      cnt[j] = 0;
    }
    dp[v][s[v - 1] - 'a'] = 1;
    cnt[s[v - 1] - 'a'] = 1;
    for (int j = 0; j < G[v].size(); ++j) {
      int to = G[v][j];
      for (int x = 0; x < 26; ++x) {
        dp[v][x] = max(dp[v][x], cnt[x] + dp[to][x]);
      }
    }
  }
  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 << endl;
  myReturn();
}
