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;

/**
 *
 * @author rrofes
 */
public class ASetOfStrings {

     
    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
       
       Scanner scan = new Scanner(System.in);
       int q = Integer.valueOf(scan.nextLine());
       String line = scan.nextLine();
       int [] hash = new int[26];
       int cuteString = 0;
       String [] b = new String[q];
       int pos=-1;
       if(q>line.length()){
           System.out.println("NO");
           System.out.close();
       }
        for (int i = 0; i < line.length(); i++) {
            char letter = line.charAt(i);
            if(hash[letter-'a']==0){
                cuteString++;
                hash[letter-'a']=1;
                if(cuteString<=q){
                    pos++;
                    b[pos]="";
                    b[pos] += letter;
                }
                else{
                    b[pos] += letter;
                }
                
            }else{
                b[pos] += letter;
                
            }
        }
        
        if(cuteString >= q){
            System.out.println("YES");
            for (int i = 0; i < b.length; i++) {
            System.out.println(b[i]);    
            }
        }else{
            System.out.println("NO");
        }
    }
    
}