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.InputStreamReader;
import java.io.IOException;
import java.util.StringTokenizer;

import java.util.Hashtable;
public class aSolution{
	private static BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
	private static StringTokenizer st;
	private static int[][] board;
	public static void main(String[] args){
		int k = Integer.parseInt(next());
		String s = next();
		String[] ans = new String[k+1];
		Hashtable<String, Integer> seen = new Hashtable<String, Integer>();
		Hashtable<Integer, String> map = new Hashtable<Integer, String>();
		String next;
		for (int i = 0; i < s.length(); i++){
			next = s.substring(i, i+1);
			if (!seen.containsKey(next)){
				seen.put(next, i);
				map.put(i, next);
			}
		}
		if (seen.size() < k){
			System.out.println("NO");
		} else {
			int count = 0;
			next = "";
			for (int i = 0; i < s.length(); i++){
				if (map.containsKey(i) && count < k){
					ans[count] = next;
					count += 1;
					next = "";
				}
				next = next + s.substring(i, i+1);
				// System.out.println(next);
			}
			ans[count] = next;
			System.out.println("YES");
			for (int i = 1; i < k+1; i++){
				System.out.println(ans[i]);
			}
		}

		
	}

	private static String next(){
		while(st == null || !st.hasMoreElements()){
			String str;
			try{
				str = br.readLine();
			} catch (IOException e){
				return null;
			}
			st = new StringTokenizer(str);
		}
		return st.nextToken();
	}

	private int gcd(int x, int y){
		if (y > x){
			return gcd(y, x);
		}
		if (y == 0){
			return x;
		}
		return gcd(y, x % y);
	}
}