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.*;
import java.util.*;
import java.math.*;

public class Div2_302_A {
	public static void main(String[] args) throws IOException {
		BufferedInputStream bis = new BufferedInputStream(System.in);
		BufferedReader br = new BufferedReader(new InputStreamReader(bis));
		int len = Integer.parseInt(br.readLine().trim());
		String input = br.readLine().trim();
		boolean [] appeared = new boolean[26];
		int distinctCnt = 0;
		for(int i = 0;i < input.length();i++) {
			int pos = input.charAt(i)-'a';
			if(!appeared[pos]) {
				appeared[pos] = true;
				distinctCnt++;
			}
		}
		if(distinctCnt < len)
			System.out.println("NO");
		else {
			System.out.println("YES");
			String curr = input.charAt(0)+"";
			appeared[input.charAt(0)-'a'] = false;
			int used = 1;
			for(int i = 1;i < input.length();i++) {
				int ch = input.charAt(i)-'a';
				if(!appeared[ch] || used == len) {
					curr += input.charAt(i); 
				} else {
					System.out.println(curr);
					used++;
					curr = input.charAt(i)+"";
					appeared[ch] = false;
				}
			}
			System.out.println(curr);
		}
	}
}