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 SetofStrings {

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int k=sc.nextInt();
        String s = sc.next();
        int len=s.length();
        List<Character> chars = new ArrayList<Character>();
        List<Integer> start = new ArrayList<Integer>();
        int i=0;
        while(k>0 && i < len)
            {
                if(!chars.contains(s.charAt(i)))
                {
                    start.add(i);
                    chars.add(s.charAt(i));
                    k--;
                }
                i++;
            }
        if(k>0)
        {
            System.out.println("NO");
            
        }else
        {
            System.out.println("YES");
            for(int j=0;j<start.size();j++)
            {
                if(j==start.size()-1)
                
                System.out.println(s.substring(start.get(j)));
                else
                    System.out.println(s.substring(start.get(j),start.get(j+1)));
            }
        }
        

    }

}