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.ArrayList;
import java.util.Scanner;
import java.util.Vector;

public class D {
    public static void main(String[] args)
    {
        D_SOLVER res = new D_SOLVER();
        res.solve();
    }
    static class D_SOLVER
    {
        int n;
        int m;
        char[] s;
        ArrayList<Integer>[] edge;
        int[][] dp;
        boolean[] visited;
        boolean[] seen;
        public void solve()
        {
            Scanner input = new Scanner(System.in);
            n = input.nextInt();
            m = input.nextInt();
            s = input.next().toCharArray();
            dp = new int[n+1][26];
            visited = new boolean[n+1];
            seen = new boolean[n+1];
            edge = new ArrayList[n+1];
            for(int i=0;i<n;i++)
            {
                edge[i] = new ArrayList<>();
            }
            for(int i=0;i<m;i++)
            {
                int x,y;
                x = input.nextInt();
                y = input.nextInt();
                edge[x-1].add(y-1);
            }
            for(int i=0;i<n;i++)
            {
                if(seen[i])
                    continue;
                if(iscycle(i))
                {
                    System.out.println(-1);
                    return;
                }
            }

            int ans = 0;
            seen = new boolean[n+1];
            for(int i=0;i<n;i++)
            {
                go(i);
                for(int j=0;j<26;j++)
                {
                    ans = Math.max(ans,dp[i][j]);
                }
            }
            System.out.println(ans);
        }
        boolean iscycle(int pos)
        {
            if(visited[pos])return true;
            if(seen[pos])return false;
            seen[pos] = true;
            visited[pos] = true;
            for(int i : edge[pos])
            {
                if(iscycle(i))
                    return true;
            }
            visited[pos] = false;
            return false;
        }
        void go(int pos)
        {
            if(seen[pos])return;
            seen[pos]=true;
            for(int i:edge[pos])
            {
                go(i);
                for(int j=0;j<26;j++)
                {
                    dp[pos][j]=Math.max(dp[pos][j],dp[i][j]);
                }
            }
            dp[pos][s[pos]-'a']++;
        }

    }
}
