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.HashSet;
import java.util.Scanner;

public class A {
	public static void main(String[] args) {
		Scanner in = new Scanner(System.in);
		int n = in.nextInt();
		String u = in.next();
		if (n > u.length()) {
			System.out.println("NO");
			return;
		}
		HashSet<Character> q = new HashSet();
		ArrayList<String> ans = new ArrayList<String>();
		for(int i = 0 ; i < u.length() ; i++){
			String k = u.charAt(i)+"";
			q.add(u.charAt(i));
			i++;
			while(i<u.length()&&(q.contains(u.charAt(i))||ans.size()==n-1))k = k + u.charAt(i++);
			if(i!=u.length())i--;
			ans.add(k);
		}
		if(ans.size()>=n){
			System.out.println("YES");
			for(int i = 0 ; i < n ; i++)System.out.println(ans.get(i));
		}else{
			System.out.println("NO");
		}
	}
}
