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.*;

public class CF302_1 {
  public static void main (String [] args) throws IOException {      
      Scanner sc = new Scanner(System.in);                                            
      PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
      
      int k = sc.nextInt();
      sc.nextLine();
      String word = sc.nextLine();
      int len = word.length();
      int minK = 1;
      ArrayList<String> parts = new ArrayList<String>();
      ArrayList<String> firstLet = new ArrayList<String>();
      for (int x = 0; x < len;) {
          int dum = 0;
          String fir = word.substring(x,x+1);
          
          for (int y = x+1; y < len;) {
              if (!word.substring(y,y+1).equals(fir) && !firstLet.contains(word.substring(y,y+1))) {
                  
                  minK++;
                  parts.add(word.substring(x,y));
                  firstLet.add(word.substring(x,x+1));
                  x = y;
                  y = len;
                  dum = 1;
              } else {
                  y++;
              }
          }
          if (dum == 0) {
              parts.add(word.substring(x,len));
              x = len;
          }
      }
      if (k > minK) {
          out.println("NO");
      } else {
          out.println("YES");
              for (int z = 0; z < k-1; z++) {
                  out.println(parts.get(z));
              }
              for (int a = k-1; a < parts.size(); a++) {
                out.print(parts.get(a));
              }
      }
      
      out.close();                                  
      System.exit(0);                              
  }
}
