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.lang.*;
import java.util.*;

public class GFG {
    static boolean[] vis; 
    // /static int[] vis;
    static int[][] dp;
    static ArrayList<ArrayList<Integer>> list;
    static boolean recStack[];
    // static int letter[];
    // static ArrayList<Integer>[] list;
    public static void main(String[] args)throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        //declaring adjacency list
        list = new ArrayList<ArrayList<Integer>>();
        
        String[] nm = br.readLine().trim().split("\\s+");
        int n = Integer.parseInt(nm[0]); //no. of nodes
        int m = Integer.parseInt(nm[1]); //no. of directed edges
        
        // list = new ArrayList[n+1];
        
        for(int i = 0;i<n+1;i++){
            list.add(i,new ArrayList<Integer>());
            // list[i] = new ArrayList<Integer>();
        }
        String s = br.readLine();
        for(int i = 0;i<m;i++){  //creating adjacency list
            String[] edges = br.readLine().trim().split("\\s+");
            int u = Integer.parseInt(edges[0]);
            int v = Integer.parseInt(edges[1]);
            list.get(u).add(v);
            // list[u].add(v);
        }
        
        // letter = new int[n+1];
        // for(int i = 1;i<=n;i++){
        //     letter[i] = s.charAt(i-1)-'a';
        // }
        
        vis = new boolean[n+1];
        dp = new int[n+1][26];
        recStack = new boolean[n+1];
        for(int i = 1;i<=n;i++){
            if(dfs(i,s)){
                System.out.println(-1);
                System.exit(0);
            }
            // if(!vis[i]){
            //     dp[v][s.charAt(i)-'a']++;
            //     SubString(list,i,vis,dp,s);
            // }
            
        }
        int answer = 0;
        for(int p = 0;p<=n;p++){
            for(int q = 0;q<26;q++){
                answer = Math.max(answer,dp[p][q]);
            }
        }
        System.out.println(answer);
        
    }
    public static boolean dfs(int vertex,String s){
        // System.out.println("dfs");
        if(recStack[vertex]){
            return true;
        }
        if(vis[vertex]){
            return false;
        }
        vis[vertex] = true;
        recStack[vertex] = true;
        for(Integer nbr: list.get(vertex)){
            // if(vis[nbr]){
            //     return true;
            // }
            // if(!vis[nbr]&&dfs(nbr,s)){
            //     return true;
            // }
            if(dfs(nbr,s)){
                return true;
            }
            for(int i = 0;i<26;i++){
                dp[vertex][i] = Math.max(dp[vertex][i],dp[nbr][i]);
            }
        }
        dp[vertex][s.charAt(vertex-1)-'a']++;
        recStack[vertex] = false;
        return false;
        
        // dp[vertex][letter[vertex]]++;

    }
}