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>
const int maxn = 300000 + 10;
using namespace std;
int indegree[maxn];
int tot[maxn][30];
bool vis[maxn];
char str[maxn];
int cou[30];
int n, m;
int ans;
struct Node {
  int to, next;
} a[maxn];
int cnt;
int head[maxn];
void insert(int x, int y) {
  a[++cnt].to = y;
  a[cnt].next = head[x];
  head[x] = cnt;
}
void dfs(int x) {
  bool flag = 1;
  for (int i = head[x]; i; i = a[i].next) {
    flag = 0;
    int y = a[i].to;
    cou[(str[y] - 'a' + 1)]++;
    dfs(y);
    cou[(str[y] - 'a' + 1)]--;
  }
  if (flag) {
    for (int i = 1; i <= 26; i++) ans = max(ans, cou[i]);
  }
}
bool topsort() {
  int cnt = 0;
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (!indegree[i]) {
      q.push(i);
      tot[i][(str[i] - 'a' + 1)]++;
      cnt++;
    }
  while (!q.empty()) {
    int k = q.front();
    q.pop();
    for (int i = head[k]; i; i = a[i].next) {
      indegree[a[i].to]--;
      for (int j = 1; j <= 26; j++)
        tot[a[i].to][j] = max(tot[a[i].to][j], tot[k][j]);
      if (!indegree[a[i].to]) {
        q.push(a[i].to);
        tot[a[i].to][(str[a[i].to] - 'a' + 1)]++;
        cnt++;
      }
    }
  }
  if (cnt < n) return 0;
  for (int i = 1; i <= n; i++)
    for (int j = 1; j <= 26; j++) ans = max(ans, tot[i][j]);
  return 1;
}
int main() {
  scanf("%d%d", &n, &m);
  scanf("%s", str + 1);
  for (int i = 1; i <= m; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    insert(x, y);
    indegree[y]++;
  }
  if (!topsort())
    printf("%d\n", -1);
  else
    printf("%d\n", ans);
  return 0;
}
