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.*;
public class Main
{
    boolean palin(String s)
    {
        for(int i=0 ; i<s.length()/2 ; i++)
        {
            if(s.charAt(i)!=s.charAt(s.length()-1-i))
             return false;
        }
        return true;
    }
	 public static void main(String args[])
    {
        Scanner in = new Scanner(System.in);
        Main ob = new Main();
        int k = in.nextInt();
      String s = in.next();
      String str[] = new String[k];
      int x=0,c=1,i=1;
      str[0]=s.substring(0,1);
      for( ; i<s.length();i++)
      {
          if(s.charAt(i)==s.charAt(i-1))
             {
                 str[x]+=String.valueOf(s.charAt(i));
             }
             else
             {
                 int flag=0;
                 for(int j=0 ; j<=x ; j++)
                 {
                     if(str[j].charAt(0)==s.charAt(i))
                     {
                         flag=1;
                         break;
                     }
                 }
                 if(flag==1)
                   str[x]+=String.valueOf(s.charAt(i));
                   else
                   {
                       if(c==k) break;
                       x++;
                       str[x]=String.valueOf(s.charAt(i));
                       c++;
                   }
             }
      }
      str[x]+=s.substring(i,s.length());
      if(c==k)
       {
           System.out.println("YES");
           for( i=0 ; i<=x ; i++)
            System.out.println(str[i]);
       }
       else
       System.out.println("NO");
    }
}
