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 Main
{
 
   static ArrayList<Integer> []graph=null;
  
  static int [] letter;
    
static int [] visited;
 
   static int [][] mem;

	public static void main(String[] args) {
 
       Scanner sc=new Scanner(System.in);
       
 int v= sc.nextInt();
      
  int e=sc.nextInt();
       
 String s=sc.next();
       
 graph=new ArrayList[v+1];
      
  for(int i=0;i<=v;i++)
     
   {
          
  graph[i]=new ArrayList<Integer>();
  
      }
        
   
     for(int i=0;i<e;i++)
        {

            int from=sc.nextInt();
     
       int to=sc.nextInt();
       
     graph[from].add(to);
        }
  
      letter=new int[v+1];
      
  for(int i=1;i<=v;i++)
   
     {
            letter[i]=s.charAt(i-1)-'a';
        }
   
     visited=new int[v+1];
      
  mem=new int[v+1][26];
   
      for(int i=1;i<=v;i++)
   
     {
          
  if(visited[i]==0 && dfs(i))
   
         {
                System.out.println("-1");
         
       System.exit(0);
            }
  
      }
        
       
 int ans=0;
       
 for(int i=0;i<=v;i++)
    
    {
          
  for(int j=0;j<26;j++)
      
      {
            
    ans=Math.max(ans,mem[i][j]);
     
       }
            
      
  }
        
      
  System.out.println(ans);
   
     
	}
	
	

	public static boolean dfs(int v)

	{
	    visited[v]=1;
	 
   for(int to:graph[v])
	  
  {
	        if(visited[to]==1)
	
        {
	            return true;

	        }
	        else if(visited[to]==0 && dfs(to))
	
        {
	            return true;
	        }
	      
  for(int i=0;i<26;i++)
	     
   {
	            mem[v][i]=Math.max(mem[v][i],mem[to][i]);
	
            
	        }
	    }      
 
	        mem[v][letter[v]]++;
	     
   visited[v]=2;
	  
      return false;
	
 
	}
	


}
