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.Scanner;

public class c302_1 {
  public static void main(String args[]){
      Scanner s=new Scanner(System.in);
      int n=s.nextInt();
      s.nextLine();
      char s1[]=s.nextLine().toCharArray();
      String s2[]=new String[n];
      int a[]=new int[200];
      int j=0;
     // System.out.println(s1[0]);
      s2[0]="";
      s2[0]+=s1[0];
      a[s1[0]]++;
      for(int i=1;i<s1.length;i++){ 
         if(j==n-1){
             for(;i<s1.length;i++)
               s2[j]+=s1[i];
             break;
         }
         if(s1[i]==s1[i-1] || a[s1[i]]!=0){
             s2[j]+=s1[i];
         }
         else{
             ++j;
             s2[j]="";
             s2[j]+=s1[i];
             a[s1[i]]++;
         }
      }
      if(s2[n-1]!=null){
          System.out.println("YES");
          for(int i=0;i<n;i++){
              System.out.println(s2[i]);
          }
          return;
      }
      else
          System.out.println("NO");
  }
}
