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.io.*;
import java.util.*;
import java.text.*;
import java.math.*;
import java.util.regex.*;

public class Solution {


   
 public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int k=sc.nextInt();
     sc.nextLine();
     int tmp=k;
        String s=sc.nextLine();
        boolean bool[]=new boolean[26];
        for(int i=0;i<s.length();i++)
        {
            if(!bool[s.charAt(i)-'a'])
            {
                bool[s.charAt(i)-'a']=true;
                k--;
            }
            
        }
     if(k>0)
     {
         System.out.println("NO");
         return;
     }
     k=tmp;
     ArrayList<Integer> index=new ArrayList();
     HashSet<Integer> set=new HashSet();
     for(int i=0;i<s.length();i++)
     {
         if(bool[s.charAt(i)-'a']&&!set.contains(s.charAt(i)-'a'))
         {
             index.add(i);
         }
         set.add(s.charAt(i)-'a');
     }
     if(index.size()==1)
     {
         System.out.println("YES");
         System.out.println(s);
         return;
     }
     System.out.println("YES");
     index.add(s.length());
     for(int i=0;i<index.size()-1;i++)
     {
         int st=index.get(i);
         int en=index.get(i+1);
         if(k==1)
         {
             System.out.println(s.substring(st));
             return;
         }
         System.out.println(s.substring(st,en));
         k--;
     }
            
   }
}
