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 cf544a {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        short k=in.nextShort();
        String q=in.next();
        if(k>q.length()){
            System.out.println("NO");
        }else {
            HashSet<Character> test = new HashSet<>();
            ArrayList<String> miniStr = new ArrayList<>();
            char current = '.';
            short start=0;
            for (short s=0;s<q.length();s++) {
                if (!test.contains(q.charAt(s))) {
                    if (current != '.' && k > 0) {
                        k--;
                        if (k == 0) {
                            miniStr.add(q.substring(start, q.length()));
                            start=(short)(q.length());
                        } else {
                            miniStr.add(q.substring(start, s));
                        }

                    }
                    if(start!=q.length())start = s;
                    current = q.charAt(s);
                    test.add(current);
                }
            }
            if (start!=q.length()){
                miniStr.add(q.substring(start, q.length()));
                k--;
            }
            if(k==0){
                System.out.println("YES");
                for (String s : miniStr) {
                    System.out.println(s);
                }
            }else{
                System.out.println("NO");
            }
        }
        in.close();
    }
}
