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 round_302;

import java.util.Arrays;
import java.util.Scanner;

// Set of Strings
public class ID_544_A {
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);

		int k = sc.nextInt();
		String q = sc.next();
		char[] a = new char[q.length()], 
				reserv = new char[q.length()];
		int[] intervals = new int[k];
		Arrays.fill(reserv, ' ');
		int distincts = 0;

		for (int i = 0; i < a.length; i++)
			a[i] = q.charAt(i);

		for (int i = 0; i < a.length; i++) {
			if (notIn(a[i], reserv)) { 
				
				if (distincts < k) { 
					intervals[distincts] = i;
					reserv[i] = a[i];
					// System.out.println("inter["+distincts+"]:" 
					//		+ intervals[distincts]); 
				}
				
				distincts++;
			}
		}

		// System.out.println("dist: " + distincts);

		if (distincts >= k) {
			System.out.println("YES");
			for (int i = 0; i < k; i++) {
				if (i == k - 1)
					System.out.println(q.substring(intervals[i]));
				else 
					System.out.println(q.substring(intervals[i], intervals[i+1]));
			}

		} else
			System.out.println("NO");

		sc.close();
	}

	public static boolean notIn(char c, char[] a) {

		for (int i = 0; i < a.length; i++)
			if (a[i] == c) return false;
		return true;
	}
}
