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.*;
import java.lang.*;




public class A302 {

    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        int k = in.nextInt();
        String s = in.next().toLowerCase().trim();
        
        Queue<String> ses = new LinkedList<String>();
        Set<Character> ces = new HashSet<Character>();
        ces.add(s.charAt(0));
        
        int start = 0;
        int end = 0;
        while (ses.size() < k && end < s.length()) {
            ++end;
            while (end < s.length() && ces.contains(s.charAt(end))) {
                ++end;
                if (end == s.length()) {
                    break;
                }
            }
            if (ses.size() == k-1) {
                end = s.length();
            }
            ses.add(s.substring(start, end));
            if (end == s.length()) {
                break;
            }
            ces.add(s.charAt(end));
            start = end;
        }
        
        if (ses.size() < k) {
            System.out.println("NO");
        } else {
            System.out.println("YES");
            for (String bit : ses)
                System.out.println(bit);
        }
    }
    
}
