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) {
        Scanner in = new Scanner(System.in);
        int k = in.nextInt();
        int initk = k;
        String s = in.next();
//        Map<String, Boolean> firstLetters = new HashMap<String, Boolean>(26);
        Set<Character> presentChars = new HashSet<Character>(26);
        List<String> strings = new ArrayList<String>(100);

        if(k == 1){
            p("YES");
            p(s);
            return;
        }

        if(k > s.length()) {
            p("NO");
            return;
        }

        StringBuilder strToAdd = new StringBuilder();
        boolean addRest = false;
        for(Character c: s.toCharArray()){
            if(addRest){
                strToAdd.append(c);
                continue;
            }
            if(strToAdd.length() == 0){
                strToAdd.append(c);
                presentChars.add(c);
            } else {
                if(presentChars.contains(c)){
                    strToAdd.append(c);
                    presentChars.add(c);
                } else {
                    k -= 1;
                    if(k == 0) {
                        strToAdd.append(c);
                        addRest = true;
                    } else {
                        strings.add(strToAdd.toString());
                        strToAdd = new StringBuilder();
                        strToAdd.append(c);
                        presentChars.add(c);
                    }
                }
            }
        }
        if (strToAdd.length() > 0) {
            strings.add(strToAdd.toString());
        }
        if(strings.size() < initk){
            p("NO");
            return;
        } else {
            p("YES");

            for(String st: strings){
                p(st);
            }
        }


    }

    public static void p(Object str){
        System.out.println(str);
    }
}
