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.
import java.util.*;

public class Main{
    public static void main(String args[]){
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        sc.nextLine();
        String a;
        a = sc.nextLine();
        
        int X[] = new int[m], Y[] = new int[m];
        for(int i = 0 ; i < m ; ++i)
        {
            X[i]=sc.nextInt();
            Y[i]=sc.nextInt();
        }
        System.out.println(beauty(n,m,a,X,Y));
    }

    public static int beauty(int n, int m, String a, int X[], int Y[]) {
        char s[] = a.toCharArray();
        int x[] = new int[m + 1];
        int y[] = new int[m + 1];

        for (int i = 0; i < m; i++) {
            x[i] = X[i] - 1;
            y[i] = Y[i] - 1;
        }
        ArrayList<TreeSet<Integer>> adj = new ArrayList<>();
        int[] incoming = new int[n];
        for (int i = 0; i < n; i++)
            adj.add(new TreeSet<Integer>());
        for (int i = 0; i < m; i++) {

            if (!adj.get(x[i]).contains(y[i])) {
                adj.get(x[i]).add(y[i]);
                incoming[y[i]]++;
            }
        }
        int[][] dp = new int[n][26];
        for (int i = 0; i < n; i++)
            dp[i][s[i] - 'a'] = 1;

        int best = Integer.MIN_VALUE;
        boolean[] visited = new boolean[n];

        ArrayDeque<Integer> que = new ArrayDeque<Integer>();

        for (int i = 0; i < n; i++)
            if (incoming[i] == 0)
                que.add(i);

        while (!que.isEmpty()) {
            int cur = que.poll();

            visited[cur] = true;

            Iterator<Integer> i = adj.get(cur).iterator();
            while (i.hasNext()) {
                int next = i.next();
                incoming[next]--;

                for (int j = 0; j < 26; j++) {
                    dp[next][j] = Math.max(dp[next][j], dp[cur][j] + ((j == (s[next] - 'a')) ? 1 : 0));
                    best = Math.max(best, dp[next][j]);
                }

                if (incoming[next] == 0)
                    que.add(next);
            }
        }

        boolean cycle = false;
        for (int i = 0; i < n; i++) {
            if (!visited[i]) {
                cycle = true;
            }
        }

        if (!cycle)
            return best;
        else
            return -1;

    }

}


