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.Scanner;


public class A
{

	public static void main(String[] args)
	{
		Scanner in = new Scanner(System.in);
		
		int k = in.nextInt(); 
		in.nextLine();
		String q = in.nextLine();
		
		in.close();
		
		int count[] = new int[256];
		
		for (int i = 0; i < q.length(); i++)
		{
			count[q.charAt(i)]++;
		}
		
		int c = 0; 
		for (int i = 'a'; i <= 'z'; i++)
		{
			if (count[i] > 0)
				c++;
		}
		
		if (c < k) //Not enough unique characters to make seperate strings
		{
			System.out.println("NO");
			return;
		}
		else
		{
			System.out.println("YES");
		}
		
		int prev = 0;
		int soln = 1;
		for (int i = 0; i < q.length(); i++)
		{
			if ((soln - 1) == k)
				break;
			
			if (count[q.charAt(i)] > 0)
			{
				if (i != 0)
					System.out.println(q.substring(prev, i));
				prev = i;
				soln++;
				count[q.charAt(i)] = 0;
			}
		}
		
		System.out.println(q.substring(prev));
	}

}
