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 class CF544A
{
    public static void main(String[] args)throws IOException
    {
        BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
        int k=Integer.parseInt(br.readLine());
        String str=br.readLine();
        int[] alp=new int[26];
        Arrays.fill(alp,-1);
        ArrayList<String> al=new ArrayList<>();
        if(str.length()<k)
        {
            System.out.println("NO");
        }
        else if(k==1)
        {
            System.out.println("YES");
            System.out.println(str);
        }
        else
        {   
            int si=0;
            int ei=1;
            alp[str.charAt(si)-'a']=1;
            for(int i=1;i<str.length();i++)
            {
                if(alp[str.charAt(i)-'a']==-1&&al.size()<k-1)
                {
                    alp[str.charAt(i)-'a']=1;
                    al.add(str.substring(si,i));
                    si=i;
                    ei=i+1;
                }
                if(alp[str.charAt(i)-'a']==-1&&al.size()==k-1)
                {
                    alp[str.charAt(i)-'a']=1;
                    al.add(str.substring(si,str.length()));
                    break;
                }
                if(i==str.length()-1)
                {
                    al.add(str.substring(si,str.length()));
                }
            }
            if(al.size()!=k)
            {
                System.out.println("NO");
            }
            else if(al.size()==k)
            {
                System.out.println("YES");
                for(String s:al)
                {
                    System.out.println(s);
                }
            }
        }
    }
}