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 MAXN = 3e5 + 5;
const long long MOD = 1e9 + 7;
int N, M, K, Q, T;
vector<int> graph[MAXN];
vector<int> reverse_graph[MAXN];
char str[MAXN];
stack<int> stk;
int vis[MAXN];
int outOfstack[MAXN];
int dp[MAXN][30];
void dfs(int u) {
  vis[u] = true;
  for (int i = 0; i < graph[u].size(); ++i) {
    int v = graph[u][i];
    if (vis[v]) continue;
    dfs(v);
  }
  stk.push(u);
}
bool dfs2(int u) {
  if (!outOfstack[u]) return false;
  vis[u] = true;
  for (int i = 0; i < reverse_graph[u].size(); ++i) {
    int v = reverse_graph[u][i];
    if (vis[v]) continue;
    if (!dfs2(v)) return false;
  }
  return true;
}
bool check() {
  for (int i = 1; i <= N; ++i) {
    if (vis[i]) continue;
    dfs(i);
  }
  for (int i = 1; i <= N; ++i) vis[i] = false;
  bool flag = true;
  while (!stk.empty()) {
    int u = stk.top();
    stk.pop();
    outOfstack[u] = true;
    if (vis[u]) continue;
    flag = dfs2(u);
    if (!flag) return flag;
  }
  return flag;
}
int go(int u, int cnum) {
  int& ret = dp[u][cnum];
  if (ret != -1) return ret;
  int val = str[u] == 'a' + cnum ? 1 : 0;
  ret = str[u] == 'a' + cnum ? 1 : 0;
  for (int i = 0; i < graph[u].size(); ++i) {
    int v = graph[u][i];
    ret = max(ret, go(v, cnum) + val);
  }
  return ret;
}
int main() {
  int i, j, u, v;
  scanf("%d%d%s", &N, &M, str + 1);
  ;
  bool flag = true;
  for (i = 1; i <= M; ++i) {
    scanf("%d%d", &u, &v);
    graph[u].push_back(v);
    reverse_graph[v].push_back(u);
    if (u == v) flag = false;
  }
  if (!flag) {
    printf("-1");
    return 0;
  }
  flag = check();
  if (!flag)
    printf("-1");
  else {
    int ans = 0;
    for (j = 0; j < 26; ++j) {
      for (i = 1; i <= N; ++i) dp[i][j] = -1;
      int ret = 0;
      for (i = 1; i <= N; ++i) {
        ret = max(ret, go(i, j));
      }
      ans = max(ans, ret);
    }
    printf("%d", ans);
  }
  return 0;
}
