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.*;
import java.lang.*;

public class sol{

	public static void main(String args[]){
		Scanner in = new Scanner(System.in);
		int k = in.nextInt();
		String s = in.next();
		if(k>s.length()) {System.out.println("NO");return;}
		int cnt[]=new int[26];
		for(int i=0;i<s.length();i++){
			cnt[s.charAt(i)-97]++;
		}
		int sum=0;
		for(int i=0;i<26;i++){
			if(cnt[i]>0) sum++;
		}
		if(sum<k) {
			System.out.println("NO");
			return;
		}
		System.out.println("YES");
		int prev=0;
		int count = 0;
		boolean arr[] = new boolean[26];
		arr[s.charAt(0)-97]=true;
		for(int i=0;i<s.length();i++){
			if(!arr[s.charAt(i)-97]){
				if(count==k-1) break;
				System.out.println(s.substring(prev,i));
				prev=i;
				arr[s.charAt(i)-97]=!arr[s.charAt(i)-97];
				count++;
			}
		}
		System.out.print(s.substring(prev,s.length()));
		return;
	}
	
	
}