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 N = 312345;
int n, m;
char used[N];
int degree[N];
char letter[N];
stack<int> st;
set<int> g[N], t[N];
int dp[N][26];
void dfs0(const int i) {
  used[i] = true;
  for (auto j : g[i])
    if (!used[j]) dfs0(j);
  st.push(i);
}
int dfs1(const int i) {
  int ans = 1;
  used[i] = true;
  for (auto j : t[i])
    if (!used[j]) ans += dfs1(j);
  return ans;
}
bool valid() {
  memset(used, false, sizeof used);
  for (int i = 0; i < n; ++i)
    if (!used[i]) dfs0(i);
  memset(used, false, sizeof used);
  for (; st.size(); st.pop())
    if (dfs1(st.top()) > 1) return false;
  return true;
}
int solve(const int i, const int kappa) {
  if (dp[i][kappa] != -1) return dp[i][kappa];
  int ans = 0;
  for (auto j : g[i]) ans = max(ans, solve(j, kappa));
  return dp[i][kappa] = ans + (letter[i] - 'a' == kappa);
}
int main() {
  scanf("%d %d", &n, &m);
  while (getchar() != '\n')
    ;
  for (int i = 0; i < n; ++i) scanf("%c", &letter[i]);
  while (m-- > 0) {
    int x, y;
    scanf("%d %d", &x, &y), --x, --y;
    g[x].insert(y), t[y].insert(x), ++degree[y];
    if (x == y) return !printf("%d\n", -1);
  }
  if (!valid()) return !printf("%d\n", -1);
  int ans = 0;
  memset(dp, -1, sizeof dp);
  for (int kappa = 'a'; kappa <= 'z'; ++kappa)
    for (int i = 0; i < n; ++i) ans = max(ans, solve(i, kappa - 'a'));
  return !printf("%d\n", ans);
}
