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.Scanner;

public class Main {

	public void solve() {
		Scanner sc = new Scanner(System.in);
		int k = sc.nextInt();
		char[] string = sc.next().toCharArray();
		boolean[] used = new boolean[26];
		ArrayList<String> ans = new ArrayList<>();

		StringBuilder sb = new StringBuilder();
		for (int i = 0; i < string.length; i++) {
			if (!used[string[i] - 'a']) {
				if (sb.length() > 0) {
					ans.add(sb.toString());
					sb.setLength(0);
				}
				used[string[i] - 'a'] = true;
			}
			sb.append(string[i]);
		}
		if (sb.length() > 0) {
			ans.add(sb.toString());
		}
		if (ans.size() < k) {
			System.out.println("NO");
			return;
		}
		System.out.println("YES");
		for (int i = 0; i < ans.size(); i++) {
			if (i < k - 1) {
				System.out.println(ans.get(i));
			} else {
				System.out.print(ans.get(i));
			}
		}
		System.out.println();

	}

	public static void main(String[] args) {
		new Main().solve();
	}

}