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.util.*;
public class Main {
    public static void main(String args[])
    {
        FastReader input=new FastReader();
        PrintWriter out=new PrintWriter(System.out);
        int T=1;
        while(T-->0)
        {
            int n=input.nextInt();
            int m=input.nextInt();
            String s=input.next();
            ArrayList<Integer> adj[]=new ArrayList[n+1];
            char ch[]=new char[n+1];
            for(int i=1;i<=n;i++)
            {
                adj[i]=new ArrayList<>();
                ch[i]=s.charAt(i-1);
            }
            int deg[]=new int[n+1];
            for(int i=0;i<m;i++)
            {
                int u=input.nextInt();
                int v=input.nextInt();
                deg[v]++;
                adj[u].add(v);
            }
            Queue<Integer> q=new LinkedList<>();
            int dp[][]=new int[n+1][26];
            for(int i=1;i<=n;i++)
            {
                if(deg[i]==0)
                {
                    q.add(i);
                }
                dp[i][(int)ch[i]-97]++;
            }
            int count=0;
            while(!q.isEmpty())
            {
                int u=q.poll();
                count++;
                for(int i=0;i<adj[u].size();i++)
                {
                    int v=adj[u].get(i);
                    for(int j=0;j<26;j++)
                    {
                        dp[v][j]=Math.max(dp[v][j],dp[u][j]+(ch[v]-97==j?1:0));
                    }
                    deg[v]--;
                    if(deg[v]==0)
                    {
                        q.add(v);
                    }
                }
            }
            if(count!=n)
            {
                out.println(-1);
            }
            else
            {
                int max=0;
                for(int i=1;i<=n;i++)
                {
                    for(int j=0;j<26;j++)
                    {
                        max=Math.max(max,dp[i][j]);
                    }
                }
                out.println(max);
            }
        }
        out.close();
    }
    static class FastReader
    {
        BufferedReader br;
        StringTokenizer st;

        public FastReader()
        {
            br = new BufferedReader(new
                    InputStreamReader(System.in));
        }

        String next()
        {
            while (st == null || !st.hasMoreElements())
            {
                try
                {
                    st = new StringTokenizer(br.readLine());
                }
                catch (IOException e)
                {
                    e.printStackTrace();
                }
            }
            return st.nextToken();
        }

        int nextInt()
        {
            return Integer.parseInt(next());
        }

        long nextLong()
        {

            return Long.parseLong(next());
        }

        double nextDouble()
        {

            return Double.parseDouble(next());
        }

        String nextLine()
        {
            String str = "";
            try
            {
                str = br.readLine();
            }
            catch (IOException e)
            {
                e.printStackTrace();
            }
            return str;
        }
    }
}