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 mod = 1e9 + 7;
const int maxn = 3e5 + 5;
const long long inf = 0x3f3f3f3f;
using namespace std;
int n, m;
int ru[maxn];
int ans;
int cnt[26];
int tmp[26];
int dp[maxn][26];
vector<int> g[maxn];
vector<int> qi;
char a[maxn];
bool vis[maxn];
bool judge() {
  queue<int> q;
  int sum = 0;
  for (int i = 1; i <= n; i++) {
    if (ru[i] == 0) q.push(i), sum++;
  }
  while (!q.empty()) {
    int s = q.front();
    q.pop();
    for (int i = 0; i < g[s].size(); i++) {
      ru[g[s][i]]--;
      if (ru[g[s][i]] == 0) sum++, q.push(g[s][i]);
    }
  }
  return sum == n;
}
void dfs(int v) {
  if (vis[v]) {
    for (int i = 0; i < 26; i++) {
      cnt[i] = max(cnt[i], dp[v][i] + tmp[i]);
    }
    return;
  }
  vis[v] = true;
  if (g[v].size() == 0) {
    tmp[a[v] - 'a']++;
    dp[v][a[v] - 'a']++;
    for (int i = 0; i < 26; i++) {
      cnt[i] = max(cnt[i], tmp[i]);
    }
    tmp[a[v] - 'a']--;
    return;
  }
  for (int i = 0; i < g[v].size(); i++) {
    tmp[a[v] - 'a']++;
    dfs(g[v][i]);
    for (int j = 0; j < 26; j++) {
      if (j == a[v] - 'a')
        dp[v][j] = max(dp[v][j], dp[g[v][i]][j] + 1);
      else
        dp[v][j] = max(dp[v][j], dp[g[v][i]][j]);
    }
    tmp[a[v] - 'a']--;
  }
}
int main() {
  std::ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n >> m;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  while (m--) {
    int x, y;
    cin >> x >> y;
    ru[y]++;
    g[x].push_back(y);
  }
  for (int i = 1; i <= n; i++) {
    if (ru[i] == 0) qi.push_back(i);
  }
  if (!judge()) {
    cout << -1 << endl;
    return 0;
  }
  for (int j = 0; j < qi.size(); j++) {
    dfs(qi[j]);
  }
  for (int i = 0; i < qi.size(); i++) {
    for (int j = 0; j < 26; j++) {
      ans = max(ans, dp[qi[i]][j]);
    }
  }
  cout << ans << endl;
  return 0;
}
