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 = (int)3e5;
int n, timer, top;
vector<int> G[N + 5];
int dfsn[N + 5], pila[N + 5], inpila[N + 5], comp[N + 5];
int dfs(int u) {
  int low = dfsn[u] = ++timer;
  inpila[pila[top++] = u] = 1;
  for (int v : G[u]) {
    if (dfsn[v] == 0)
      low = min(low, dfs(v));
    else if (inpila[v])
      low = min(low, dfsn[v]);
  }
  if (low == dfsn[u]) {
    int fin;
    do {
      fin = pila[--top];
      inpila[fin] = 0;
      comp[fin] = u;
    } while (fin != u);
  }
  return low;
}
void SCC() {
  memset(dfsn, 0, sizeof(dfsn));
  top = timer = 0;
  for (int i = 0; i < n; i++)
    if (!dfsn[i]) dfs(i);
}
void add(int u, int v) { G[u].push_back(v); }
bool used[N][26];
int memo[N][26];
string letters;
int dp(int u, int target) {
  if (used[u][target]) {
    return memo[u][target];
  }
  used[u][target] = true;
  int &ans = memo[u][target] = (letters[u] - 'a' == target);
  for (int i = 0; i < ((int)(G[u]).size()); i++) {
    int v = G[u][i];
    ans = max(ans, (letters[u] - 'a' == target) + dp(v, target));
  }
  return ans;
}
int main() {
  ios_base ::sync_with_stdio(false);
  int m;
  while (cin >> n >> m) {
    cin >> letters;
    for (int i = 0; i < n; i++) {
      G[i].clear();
    }
    vector<int> indeg(n);
    bool hasCycle = false;
    for (int i = 0; i < m; i++) {
      int u, v;
      cin >> u >> v;
      u--, v--;
      if (u == v) {
        hasCycle = true;
      }
      add(u, v);
      indeg[v]++;
    }
    SCC();
    for (int i = 0; i < n; i++) {
      if (comp[i] != i) {
        hasCycle = true;
      }
    }
    if (hasCycle) {
      cout << -1 << '\n';
      continue;
    }
    memset(used, false, sizeof(used));
    int ans = 0;
    for (int letter = 0; letter < 26; letter++) {
      for (int u = 0; u < n; u++) {
        if (indeg[u] == 0) {
          ans = max(ans, dp(u, letter));
        }
      }
    }
    cout << ans << '\n';
  }
}
