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 GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,sse3,sse4,popcnt,abm,mmx")
const int N = (int)5e5 + 7;
const int inf = (int)1e9 + 7;
const int mod = (int)1e9 + 7;
const long long linf = (long long)1e18 + 7;
const int dx[] = {-1, 0, 1, 0, 1, -1, -1, 1};
const int dy[] = {0, 1, 0, -1, 1, -1, 1, -1};
using namespace std;
int n, m;
int a[N];
vector<int> g[N], gr[N];
int was[N];
int dp[N][26];
vector<int> cmp;
void dfs(int v) {
  was[v] = 1;
  for (auto to : g[v]) {
    if (!was[to]) {
      dfs(to);
    } else if (was[to] == 1) {
      cout << "-1", exit(0);
    }
  }
  was[v] = 2;
}
void dfs1(int v) {
  was[v] = 1;
  for (auto to : g[v]) {
    if (!was[to]) dfs1(to);
  }
  cmp.push_back(v);
}
int calc(int v, int x) {
  if (~dp[v][x]) return dp[v][x];
  int res = 0;
  for (auto to : g[v]) {
    res = max(res, calc(to, x));
  }
  return dp[v][x] = res + (a[v] == x);
}
void solve() {
  for (int i = 1; i <= n; i++) {
    if (!was[i]) dfs(i);
  }
  memset(was, 0, sizeof(was));
  for (int i = 1; i <= n; i++) {
    if (!was[i]) {
      dfs1(i);
    }
  }
  int ans = 0;
  reverse(cmp.begin(), cmp.end());
  memset(dp, -1, sizeof(dp));
  for (int i = 0; i < 26; i++) {
    for (auto it : cmp) {
      ans = max(ans, calc(it, i));
    }
  }
  cout << ans;
}
int main() {
  ios_base ::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    char x;
    cin >> x;
    a[i] = x - 'a';
  }
  for (int i = 1; i <= m; i++) {
    int v, u;
    cin >> v >> u;
    if (v == u) cout << -1, exit(0);
    g[v].push_back(u);
  }
  solve();
  exit(0);
}
