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 Main{
	public static void main(String args[]){
		int n;
		
		Scanner ss = new Scanner(System.in);
		n=ss.nextInt();
		String s = ss.next();
		int a[] = new int[27];
		int i;
		int  b[] = new int[27];
		for(i=0;i<27;i++){
			b[i]=0;
			a[i]=0;
		}
		for(i=0;i<s.length();i++){
			a[s.charAt(i)-'a']=1;
		}
		int sum=0;
		int cot=0;
		for(i=0;i<27;i++)
			if(a[i]!=0)
				sum++;
		if(sum>=n){
			System.out.println("YES");
			for(i=0;i<s.length();i++){
				if(b[s.charAt(i)-'a']==0){
					if(i==0)
						System.out.print(s.charAt(i));
					else
						System.out.print("\n"+s.charAt(i));
					b[s.charAt(i)-'a']=1;
					cot++;
				}
				else
					System.out.print(s.charAt(i));
				if(cot==n)
					break;
			}
			//System.out.println(i);
			for(++i;i<s.length();i++)
				System.out.print(s.charAt(i));
			System.out.println();
		}
		else
			System.out.println("NO");
	}
}