You are given a string q. A sequence of k strings s1, s2, ..., sk is called beautiful, if the concatenation of these strings is string q (formally, s1 + s2 + ... + sk = q) and the first characters of these strings are distinct.

Find any beautiful sequence of strings or determine that the beautiful sequence doesn't exist.

Input

The first line contains a positive integer k (1 ≤ k ≤ 26) — the number of strings that should be in a beautiful sequence. 

The second line contains string q, consisting of lowercase Latin letters. The length of the string is within range from 1 to 100, inclusive.

Output

If such sequence doesn't exist, then print in a single line "NO" (without the quotes). Otherwise, print in the first line "YES" (without the quotes) and in the next k lines print the beautiful sequence of strings s1, s2, ..., sk.

If there are multiple possible answers, print any of them.

Examples

Input

1
abca


Output

YES
abca


Input

2
aaacas


Output

YES
aaa
cas


Input

4
abc


Output

NO

Note

In the second sample there are two possible answers: {"aaaca", "s"} and {"aaa", "cas"}.

import java.util.HashSet;
import java.util.Scanner;


public class set_of_string {

    public static void main(String[] args) {
        try
        {
            Scanner s=new Scanner(System.in);
            int n=s.nextInt();
            String str=s.next();
            int count=0;
            HashSet<Character> h=new HashSet<Character>();
            if(n==1)
            {
                System.out.println("YES");
                System.out.println(str);
            }
            else
            {
               
               for(int i=0;i<str.length();i++)
               {
                 h.add(str.charAt(i));
               }
               if(h.size()<n)
               {
                   System.out.println("NO");
                   return;
               }
               h.clear();
                System.out.println("YES");
               for(int i=0;i<str.length();i++)
               {
                   
                   if(!h.contains(str.charAt(i)))
                   {
                       h.add(str.charAt(i));
                       System.out.println();
                       
                       count++;
                       if(count==n)
                       {
                           System.out.println(str.substring(i));    
                           return;
                       }
                       System.out.print(str.charAt(i));
                   }
                   else
                   {
                       System.out.print(str.charAt(i));
                   }
                   
               }
            }
        }
        catch(Exception e)
        {
            System.out.println("ERROR IS :"+e.getMessage());
        }
    }
    
}
