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.*;
public class SetString {

	public static void main(String[] args) {
		// TODO Auto-generated method stub
		Scanner sc= new Scanner(System.in);
		int k=sc.nextInt();
		String s=sc.next();
		if(k==1)
		{
			System.out.println("YES");
			System.out.println(s);
		}
		else if(k>s.length())
			System.out.println("NO");
		else 
		{
			int f[]=new int[26];
			int c=0;
			for(int i=0;i<s.length();i++)
				f[s.charAt(i)-'a']++;
			for(int i=0;i<26;i++)
				if(f[i]>0)
					c++;
			/*if(c==s.length())
			{ 
				int i=0;
				System.out.println("YES");
				for( i=0;i<k-1;i++)
					System.out.println(s.charAt(i));
				System.out.println(s.substring(i,s.length()));
			}*/
			 if(c>=k)
			{
				String x="";
				x+=s.charAt(0);
				int count=0,l=0,i=0;
				
					System.out.println("YES");
					for(int j=0;j<s.length();j++)
					{	
						if(count==k-1)
						{
							System.out.println(s.substring(l,s.length()));
							break;
						}
						for(i=0;i<x.length();i++)
						{
							if(s.charAt(j)==x.charAt(i))
								break;
								
						}
						if(i==x.length())
						{
							//System.out.println(i+" "+l+" "+j);
							System.out.println(s.substring(l,j));
							l=j;
							count++;
							x+=s.charAt(j);
							j--;
						}
							
					}
				
				
			}
			else
				System.out.println("NO");
		}
	}

}
