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.Arrays;
import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		String[] x = new String[]{"a","b","c","d","e","f","g","h","i","j","k","l","m","n","o","p","q","r","s","t","u","v","w","x","y","z"};
		int c = in.nextInt();
		String s = in.next();
		int[] index = new int[c];
		int j = 0;
		for(int i=0;i<26;i++)
		{
			if(s.indexOf(x[i])!=-1)
			{
				index[j] = s.indexOf(x[i]);
				j++;
				if(j==c)
					break;
			}
		}
		if(j!=c)
			System.out.println("NO");
		else
		{
			System.out.println("YES");
			Arrays.sort(index);
			if(c!=1)
			{
				index[0] = 0;
			for(int i=0;i<c-1;i++)
			{
				System.out.println(s.substring(index[i], index[i+1]));
			}
			System.out.println(s.substring(index[c-1]));
			}
			else
				System.out.println(s);
		}
	}

}

    			 	  			 	 	  	  				   	