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 Solution
{
    static ArrayList<Integer> [] g;
    static int [] letter;
    static int [] visited;
    static int[][] mem;
    public static void main(String [] args)
    {
        Scanner sc=new Scanner(System.in);
        int ver=sc.nextInt();
        int edg=sc.nextInt();
        String s=sc.next();
        g=new ArrayList[ver+1];
        for(int i=0;i<=ver;i++)
        {
            g[i]=new ArrayList<Integer>();
        }
        for(int i=0;i<edg;i++)
        {
            int start=sc.nextInt();
            int end=sc.nextInt();
            g[start].add(end);
        }
        letter =new int[ver+1];
        for(int i=1;i<=ver;i++)
        {
            letter[i]=s.charAt(i-1)-'a';
        }
        visited=new int[ver+1];
        mem=new int[ver+1][26];
        for(int i=1;i<=ver;i++)
        {
            if(visited[i]==0 && check(i))
            {
                System.out.println("-1");
                System.exit(0);
            }
        }
        int ans=0;
        for(int i=1;i<=ver;i++)
        {
            for(int j=0;j<26;j++)
            {
                ans=Math.max(ans,mem[i][j]);
            }
        }
        System.out.println(ans);
    }
    public static boolean check(int vertex)
    {
        visited[vertex]=1;
        for(int to:g[vertex])
        {
            if(visited[to]==1)
                return true;
            if(visited[to]==0 && check(to))
            {
                return true;
            }
            for(int i=0;i<26;i++)
            {
                mem[vertex][i]=Math.max(mem[vertex][i],mem[to][i]);
            }
        }
        mem[vertex][letter[vertex]]++;
        visited[vertex]=2;
        return false;
    }
}