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 MAXN = 300005;
vector<int> e[MAXN];
char str[MAXN];
int vis[MAXN], cir;
void dfs1(int u) {
  cir |= (vis[u] == 1);
  if (vis[u]) return;
  vis[u] = 1;
  for (int i = 0; i < (int)e[u].size(); i++) {
    int v = e[u][i];
    dfs1(v);
  }
  vis[u] = 2;
}
int dp[MAXN];
void dfs2(int u, int sp) {
  if (vis[u]) return;
  vis[u] = 1;
  int mx = 0;
  for (int i = 0; i < (int)e[u].size(); i++) {
    int v = e[u][i];
    dfs2(v, sp);
    mx = max(mx, dp[v]);
  }
  dp[u] = mx + (str[u] - 'a' == sp);
}
int main() {
  int n, m;
  scanf("%d%d%s", &n, &m, str + 1);
  for (int i = 1; i <= m; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    e[u].push_back(v);
  }
  for (int i = 1; i <= n; i++)
    if (!vis[i]) dfs1(i);
  if (cir) return 0 * printf("-1\n");
  int res = 0;
  for (int _ = 0; _ < 26; _++) {
    for (int i = 1; i <= n; i++) dp[i] = vis[i] = 0;
    for (int i = 1; i <= n; i++)
      if (!vis[i]) dfs2(i, _);
    for (int i = 1; i <= n; i++) res = max(res, dp[i]);
  }
  return 0 * printf("%d\n", res);
}
