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.ArrayList;
import java.util.Scanner;

public class A {
    public static void main(String cd[]){
        Scanner in=new Scanner(System.in);
        int k=in.nextInt();
        String s=in.next();
        int array[]=new int[200];
        ArrayList<Integer> dif=new ArrayList<>();
        for(int i=0;i<s.length();i++){
            
            if(array[s.charAt(i)]==0){
                dif.add(i);
            }
            array[s.charAt(i)]++;
        }
        if(dif.size()<k){
            System.out.println("NO");
            return;
        }
        System.out.println("YES");
        for(int i=0;i<k;i++){
            int start=dif.get(i);
            if(i!=k-1){
                int end=dif.get(i+1);
                System.out.println(s.substring(start,end));
            }
            else{
                System.out.println(s.substring(start,s.length()));
            }
        }
    }
}
