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"}.
//package round302;

import java.util.HashSet;
import java.util.LinkedList;
import java.util.Queue;
import java.util.Scanner;

public class SetOfStrings {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int k = sc.nextInt();
		String s = sc.next();

		if (k > s.length())
			System.out.println("NO");
		else {
			Queue<String> res = isBeautiful(s, k);
			if (res != null) {
				System.out.println("YES");
				for (String w : res)
					System.out.println(w);
			} else
				System.out.println("NO");
		}
		sc.close();
	}

	private static Queue<String> isBeautiful(String s, int words) {
		HashSet<Character> chars = new HashSet<Character>();
		Queue<String> results = new LinkedList<String>();
		int oldindex = 0;

		for (int i = 0; i < s.length(); i++) {
			if (words == 1)
				break;

			if (!chars.contains(s.charAt(i))) {
				chars.add(s.charAt(i));
				while ((i < s.length() - 1)
						&& (chars.contains(s.charAt(i)) || s.charAt(i) == s
								.charAt(i + 1)))
					i++;

				results.add(s.substring(oldindex, i));
				words--;
				oldindex = i--;
			}
		}

		if (words == 1 && oldindex <= s.length() - 1
				&& !chars.contains(s.charAt(oldindex))) {
			results.add(s.substring(oldindex, s.length()));
			words--;
		}

		if (words == 0)
			return results;
		return null;
	}

}
