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"}.
// Don't place your source in a package
import java.util.*;
import java.lang.*;
import java.io.*;

// Please name your class Main
public class Main {
	public static void main (String[] args) throws java.lang.Exception {
	    Scanner in = new Scanner(System.in);
	    int n = in.nextInt();
        String s = in.next();
        Set<Character> set = new HashSet<>();
        ArrayList<Integer> v = new ArrayList<>();
        for (int i = 0; i < s.length(); i++) {
            if (set.contains(s.charAt(i))) continue;
            set.add(s.charAt(i));
            v.add(i);
            if (v.size() == n) break;
        }
        if (v.size() < n) {
            System.out.println("NO");
            return;
        }
        System.out.println("YES");
        for (int i = 0; i < v.size(); i++) {
            int from, to;
            if (i + 1 < v.size()) {
                from = v.get(i);
                to = v.get(i + 1);
            } else {
                from = v.get(i);
                to = s.length();
            }
            System.out.println(s.substring(from, to));
        }
	}
}