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.HashSet;
import java.util.Scanner;

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

		ArrayList<StringBuilder> a = new ArrayList<StringBuilder>();
		HashSet<Character> used = new HashSet<Character>();
		
		a.add(new StringBuilder(s.charAt(0) + ""));
		used.add(s.charAt(0));
		for (int i = 1; i < s.length(); i++) {
//			System.out.println(a);
			char c = s.charAt(i);
			if (a.size() == n) {
				if (a.isEmpty())
					a.add(new StringBuilder());
				a.get(a.size() - 1).append(c);
			} else {
				if (used.contains(c))
					a.get(a.size() - 1).append(c);
				else {
					a.add(new StringBuilder(c + ""));
					used.add(c);
				}
			}
			
		}

		if (a.size() == n) {
			System.out.println("YES");
			for (StringBuilder ss : a)
				System.out.println(ss);
		} else
			System.out.println("NO");
	}
}
