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;
int dx[] = {0, 0, 1, -1, -1, -1, 1, 1};
int dy[] = {1, -1, 0, 0, -1, 1, 1, -1};
template <class T>
inline T biton(T n, T pos) {
  return n | ((T)1 << pos);
}
template <class T>
inline T bitoff(T n, T pos) {
  return n & ~((T)1 << pos);
}
template <class T>
inline T ison(T n, T pos) {
  return (bool)(n & ((T)1 << pos));
}
template <class T>
inline T gcd(T a, T b) {
  while (b) {
    a %= b;
    swap(a, b);
  }
  return a;
}
template <typename T>
string NumberToString(T Number) {
  ostringstream second;
  second << Number;
  return second.str();
}
inline int nxt() {
  int aaa;
  scanf("%d", &aaa);
  return aaa;
}
inline long long int lxt() {
  long long int aaa;
  scanf("%lld", &aaa);
  return aaa;
}
inline double dxt() {
  double aaa;
  scanf("%lf", &aaa);
  return aaa;
}
template <class T>
inline T bigmod(T p, T e, T m) {
  T ret = 1;
  for (; e > 0; e >>= 1) {
    if (e & 1) ret = (ret * p) % m;
    p = (p * p) % m;
  }
  return (T)ret;
}
int ar[300010];
char s[300010];
vector<int> adj[300010];
bool cycle = 0;
int color[300010];
vector<int> top;
int res = 0;
void dfs(int u) {
  color[u] = 1;
  for (int i = 0; i < adj[u].size(); i++) {
    int v = adj[u][i];
    if (color[v] == 0) {
      dfs(v);
    } else if (color[v] == 1) {
      cycle = 1;
      return;
    }
  }
  color[u] = 2;
  top.push_back(u);
}
int mark[30];
char now;
map<pair<int, int>, char> mp;
int dp[300010][30];
int vis[300010][30];
int again(int u, int c) {
  if (vis[u][c] == 1) return 0;
  if (vis[u][c] == 2) return dp[u][c];
  vis[u][c] = 1;
  int res = 0;
  for (auto v : adj[u]) {
    res = max(res, again(v, c) + ((s[v - 1] - 'a') == c));
  }
  vis[u][c] = 2;
  return dp[u][c] = res;
}
int main() {
  int n = nxt();
  int m = nxt();
  scanf("%s", s);
  for (int i = 0; i < m; i++) {
    int a = nxt();
    int b = nxt();
    if (mp.count(make_pair(a, b))) continue;
    mp[make_pair(a, b)] = 1;
    adj[a].push_back(b);
  }
  for (int i = 1; i <= n; i++) {
    if (color[i] == 0) {
      dfs(i);
      if (cycle) {
        cout << -1 << endl;
        return 0;
      }
    }
  }
  reverse(top.begin(), top.end());
  int ans = 0;
  for (int i = 0; i <= 25; i++)
    for (auto j : top) {
      ans = max(ans, again(j, i) + ((s[j - 1] - 'a') == i));
    };
  cout << ans << endl;
  return 0;
}
