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.io.*;
import java.text.DecimalFormat;
import java.util.*;
public class Main
{
    static StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out=new PrintWriter(new OutputStreamWriter(System.out));
   static int nextInt()throws IOException{
        in.nextToken();
        return (int)in.nval;
    }
    static double nextDouble()throws IOException{
        in.nextToken();
        return in.nval;
    }
    static String next() throws IOException {
       in.nextToken();
       return in.sval;
    }
    static long nextLong() throws IOException{
        in.nextToken();
        return (long)in.nval;
    }
    static Queue<Integer> q=new LinkedList<Integer>();
    static DecimalFormat df=new DecimalFormat("0.000");
    static int n,m,k,s,op,ed,maxans;
    static char[] map=new char[300050];
    static int[] p=new int[300050];
    static int[] way=new int[300050];
    static int[] lst=new int[300050];
    static int[] rd=new int[300050];
    static int[] l=new int[300050];
    static int[][] ans=new int[300050][26];
    public static void main(String args[])throws IOException{

        n=nextInt();
        m=nextInt();

        String temp=next();
        map=temp.toCharArray();
        temp="";

        for(int i=1;i<=m;i++){
            k=nextInt();
            s=nextInt();
            rd[s]++;
            ed++;
            way[ed]=s;
            lst[ed]=p[k];
            p[k]=ed;
        }

        ed=0;
        for(int i=1;i<=n;i++){
            if(rd[i]==0){
                ed++;
                l[ed]=i;
            }
        }
        while(op<ed){
            op++;
            int nw=p[l[op]];
            while(nw!=0){
                rd[way[nw]]--;
                if(rd[way[nw]]==0){
                    ed++;
                    l[ed]=way[nw];
                }
                nw=lst[nw];
            }
        }
        if(ed!=n){
            out.println(-1);
            out.flush();
            return;
        }

        for(int i=1;i<=n;i++){
            ans[i][map[i-1]-'a']=1;
        }
        for(int i=1;i<=n;i++){
            int nw=p[l[i]];
            while(nw!=0){
                for(int j=0;j<26;j++) {
                    if(j==map[way[nw]-1]-'a' && ans[way[nw]][j] < ans[l[i]][j]+1) ans[way[nw]][j] = ans[l[i]][j]+1;
                    else if (ans[way[nw]][j] < ans[l[i]][j]) ans[way[nw]][j] = ans[l[i]][j];

                }
                nw=lst[nw];
            }
        }
        for(int i=1;i<=n;i++){
            for(int j=0;j<26;j++){
                if(ans[i][j]>maxans){
                    maxans=ans[i][j];
                }
            }
        }
        out.println(maxans);
        out.flush();
    }
}
