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.Arrays;
import java.util.Scanner;

public class SetOfStrings {
    public static void main(String asd[])throws Exception{
        Scanner in=new Scanner(System.in);
        int k=Integer.parseInt(in.nextLine());
        String s=in.nextLine();
        String ss[]=new String[k];
        Arrays.fill(ss,"");
        int c[]=new int[26];
        if(s.length()<k)
            System.out.println("NO");
        else if(k==1)
            System.out.println("YES\n"+s);
        else{
            c[s.charAt(0)-'a']=1;
           int kk=0,i=0;
            ss[kk]+=s.charAt(0);
            for(i=1;i<s.length()&&kk<k-1;i++)
            {
                if(c[s.charAt(i)-'a']==0)
                {
                    kk+=1;
                    ss[kk]+=s.charAt(i);
                    c[s.charAt(i)-'a']=1;
                }
                else
                    ss[kk]+=s.charAt(i);
            }
            if(i!=s.length())
            ss[kk]+=s.substring(i,s.length());
            if(kk==k-1)
            {
                System.out.println("YES");
                for(i=0;i<k;i++)
                    System.out.println(ss[i]);
            }
            else
                System.out.println("NO");
        }
    }
}
