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.List;
import java.util.Scanner;


public class main {
	
	public static void main(String[] args) {
		Scanner scanner = new Scanner(System.in);
		int k = Integer.valueOf(scanner.nextLine());
		String q = scanner.nextLine();
		
		List<String> list = new ArrayList<String>();
		List<String> used = new ArrayList<String>();
		String cur = "";
		
		for(int i = 0; i < q.length(); i++) {
			// Does the letter start a new word?
			if(!used.contains(q.charAt(i) + "")) {
				// Add old seq to list
				if(!cur.equals("")) {
					list.add(cur);
				}
				// Create a new seq
				cur = "";
				used.add(q.charAt(i) + "");
				
			}
			cur += q.charAt(i);
			
		}
		list.add(cur);
		
		if(list.size() >= k) {
			System.out.println("YES");
			for(int i = 0; i < k; i++) {
				if(i != k - 1) {
					System.out.println(list.get(i));
					
				}
				else {
					// Build last String
					String s = "";
					for(int j = i; j < list.size(); j++) {
						s += list.get(j);
						
					}
					System.out.println(s);
					
				}
				
			}
			
		}
		else {
			System.out.println("NO");
		}
		

	}
	

}
