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"}.
/**
 * Created by Paul on 5/7/2015.
 */
import java.util.Scanner;
import java.util.Vector;

public class Main {
    private static void printStr(String s, int from, int to){
        String p = s.substring(from, to);
        System.out.println(p);
    }
    public static void main(String[] args){
        try(Scanner sc = new Scanner(System.in)){
            int k = sc.nextInt();
            String q = sc.next();

            boolean[] mark = new boolean[26];

            if(q.length() < k){
                System.out.println("NO");
                return;
            }

            if(k == 1){
                System.out.println("YES");
                System.out.println(q);
                return;
            }

            Vector<Integer> segments = new Vector<>();

            for(int i = 0; i < q.length(); ++i){
                if(!mark[q.charAt(i) - 'a']){
                    mark[q.charAt(i) - 'a'] = true;
                    segments.add(i);
                }
            }

            segments.add(segments.size());
            if(segments.size() <= k){
                System.out.println("NO");
            }
            else{

                segments.subList(k, segments.size()).clear();
                segments.add(q.length());

                System.out.println("YES");
                for(int i = 1; i < segments.size(); ++i){
                    printStr(q, segments.elementAt(i - 1), segments.elementAt(i));
                }
            }
        }
        catch (Exception e){
            e.printStackTrace();
        }
    }
}
