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.ArrayList;
import java.util.HashSet;
import java.util.Scanner;
import java.util.Set;


public class CF302A {
    public static void main(String[] args){
        Scanner keyboard = new Scanner(System.in);
        
        int n = keyboard.nextInt();
        keyboard.nextLine();
        String s = keyboard.nextLine();
        Set letters = new HashSet();
        char currentLetter = s.charAt(0);
        letters.add(currentLetter);
        ArrayList<Integer> words = new ArrayList<>();
        int k = 1;
        
        for (int i = 1; i < s.length() && k < n; i++) {
            char c = s.charAt(i);
            if(!letters.contains(c)) {
                words.add(i);
                currentLetter = c;
                letters.add(c);
                k++;
            }
        }
        
        if(k < n) {
            System.out.println("NO");
            return;
        }
        
        System.out.println("YES");
        int previous = 0;
        for (int i : words) {
            System.out.println(s.substring(previous, i));
            previous = i;
        }
        System.out.println(s.substring(previous));
                
    }
}
