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.io.*;
import java.util.*;
public final class set_of_strings
{
	static Scanner sc=new Scanner(System.in);
	static PrintWriter out=new PrintWriter(System.out);
	
    public static void main(String args[]) throws Exception 
	{
		int[] cnt=new int[123];
		int k=sc.nextInt(),dis=0;
		char[] a=sc.next().toCharArray();
		List<Integer> list=new ArrayList<Integer>();
		for(int i=0;i<a.length;i++)
		{
			if(cnt[a[i]]==0)
			{
				if(list.size()<k)
				{
					list.add(i);
				}
				dis++;
				cnt[a[i]]++;
			}
		}
		if(dis>=k)
		{
			out.println("YES");
			if(k==1)
			{
				out.println(new String(a));
			}
			else
			{
				for(int i=0;i<list.size();i++)
				{
					int curr=list.get(i);
					int next;
					if(i==list.size()-1)
					{
						next=a.length;
					}
					else
					{
						next=list.get(i+1);
					}
					for(int j=curr;j<next;j++)
					{
						out.print(a[j]);
					}
					out.println("");
				}
			}
		}
		else
		{
			out.println("NO");
		}
		out.close();
	}
}