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 INF = 0x3f3f3f3f;
const int mod = 998244353;
const int N = 3e5 + 10;
int val[N];
bool vis[50];
struct node {
  int v, next;
} edge[N];
int cnt, head[N];
char s[N];
int in[N];
int dp[N][50];
void init() {
  memset(head, -1, sizeof head);
  memset(dp, 0, sizeof dp);
  cnt = 0;
  memset(vis, 0, sizeof vis);
  memset(in, 0, sizeof in);
}
void addedge(int u, int v) {
  edge[cnt].v = v;
  edge[cnt].next = head[u];
  head[u] = cnt++;
}
int n, m;
int bfs() {
  queue<int> q;
  for (int i = 1; i <= n; i++)
    if (in[i] == 0) q.push(i);
  while (!q.empty()) {
    int t = q.front();
    q.pop();
    for (int i = head[t]; ~i; i = edge[i].next) {
      int v = edge[i].v;
      for (int j = 0; j < 26; j++) {
        dp[v][j] = max(dp[v][j], dp[t][j]);
      }
      dp[v][val[t]] = max(dp[v][val[t]], dp[t][val[t]] + 1);
      if (--in[v] == 0) q.push(v);
    }
    if (head[t] == -1) dp[t][val[t]] += 1;
  }
  bool flag = 1;
  int ans = 0;
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 26; j++) ans = max(ans, dp[i][j]);
    if (in[i]) flag = 0;
  }
  if (flag) return ans;
  return -1;
}
int main() {
  while (~scanf("%d%d", &n, &m)) {
    init();
    scanf("%s", s);
    for (int i = 0; i < n; i++) {
      val[i + 1] = s[i] - 'a';
    }
    for (int i = 0; i < m; i++) {
      int u, v;
      scanf("%d%d", &u, &v);
      addedge(u, v);
      in[v]++;
    }
    printf("%d\n", bfs());
  }
  return 0;
}
