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 double eps = 1e-6;
const int inf = 0x3f3f3f3f;
const long long INF = 0x3f3f3f3f3f3f3f3f;
const int MAXN = 1e5 + 10;
const int MAXM = 1e6 + 10;
int flag;
int ans;
struct edge {
  int v, nex;
} e[300010];
int head[300010], tot;
int d[300010];
char s[300010];
int cnt[300010][30];
bool vis[300010];
void init() {
  flag = 0;
  memset((d), (0), sizeof(d));
  memset((head), (-1), sizeof(head));
  tot = 0;
  memset((cnt), (0), sizeof(cnt));
}
void add(int u, int v) {
  e[tot].v = v;
  e[tot].nex = head[u];
  head[u] = tot++;
}
int main() {
  int n, m;
  while (~scanf("%d%d", &n, &m)) {
    scanf("%s", s + 1);
    init();
    for (int i = 0; i < m; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      if (u == v) flag = 1;
      add(u, v);
      d[v]++;
    }
    if (flag) {
      cout << -1 << endl;
      continue;
    }
    queue<int> q;
    int c = 0;
    while (!q.empty()) q.pop();
    ans = -1;
    for (int i = 1; i <= n; i++) {
      if (!d[i]) {
        c++;
        q.push(i);
      }
    }
    while (!q.empty()) {
      int u = q.front();
      cnt[u][s[u] - 'a']++;
      for (int i = 0; i < 26; i++) {
        ans = max(ans, cnt[u][i]);
      }
      q.pop();
      for (int i = head[u]; i != -1; i = e[i].nex) {
        int v = e[i].v;
        for (int i = 0; i < 26; i++) {
          cnt[v][i] = max(cnt[u][i], cnt[v][i]);
        }
        d[v]--;
        if (d[v] == 0) {
          c++;
          q.push(v);
        }
      }
    }
    if (c != n)
      puts("-1");
    else
      cout << ans << endl;
  }
  return 0;
}
