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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;


public class SetOfStrings {

	public static void main(String[] args) throws IOException {
		BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
		int jml = Integer.parseInt(in.readLine());
		ArrayList <String> listWord = new ArrayList <String>();
		int [] listChar = new int [26];
		String str = in.readLine();
		
		for(int ii = 0; ii < str.length(); ii++){
			if(listChar[str.charAt(ii)-97] == 0){
				listChar[str.charAt(ii)-97] = 1;
				listWord.add(str.charAt(ii)+"");
			}else{
				String temp = listWord.get(listWord.size()-1);
				listWord.remove(listWord.size()-1);
				listWord.add(temp + str.charAt(ii));
				
			}
				
		}
		
		int count = 0;
		if(listWord.size() < jml){
			System.out.println("NO");
		}
		
		else{
			System.out.println("YES");
			for(int ii = 0; ii<jml-1; ii++){
				count = count + listWord.get(ii).length();
				System.out.println(listWord.get(ii));
			}
			System.out.println(str.substring(count));
			
		}
		
	}
}
