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 comment(linker, "/STACK:102400000,102400000")
using namespace std;
const int maxn = 300000 + 5;
const int INF = 0x3f3f3f3f;
const int P = 1000000007;
const double PI = acos(-1.0);
template <typename T>
inline T read(T &x) {
  x = 0;
  int _f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') _f |= (ch == '-'), ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x = _f ? -x : x;
}
template <class T1, class T2>
inline void gmax(T1 &a, T2 b) {
  if (b > a) a = b;
}
template <class T1, class T2>
inline void gmin(T1 &a, T2 b) {
  if (b < a) a = b;
}
bool flag;
int n, m, u, v, ret, num[maxn][26];
vector<int> G[maxn];
char s[maxn];
bool vis[maxn][2];
void umax(int &a, int b) {
  if (a < b) a = b;
}
void dfs(int u) {
  if (flag) return;
  if (vis[u][0]) return;
  vis[u][1] = true;
  for (int i = 0; i < (int)G[u].size(); i++) {
    int v = G[u][i];
    if (vis[v][1]) {
      flag = true;
      return;
    } else
      dfs(v);
    if (flag) return;
    for (int j = 0; j < 26; j++) umax(num[u][j], num[v][j]);
  }
  num[u][s[u] - 'a']++;
  vis[u][0] = true, vis[u][1] = false;
}
int main() {
  read(n), read(m);
  scanf("%s", s + 1);
  for (int i = 0; i < m; i++) {
    read(u), read(v);
    G[u].push_back(v);
  }
  for (int i = 1; i <= n; i++) {
    if (!vis[i][0]) dfs(i);
    if (flag) return puts("-1"), 0;
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 0; j < 26; j++) {
      umax(ret, num[i][j]);
    }
  }
  cout << ret << endl;
  return 0;
}
