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.ArrayList;
import java.util.Scanner;

/**
 * Created by pinguinson on 07.05.2015.
 */
public class A {
    public static void main(String[] args) {

        Scanner sc = new Scanner(System.in);
        int k = sc.nextInt();
        String s = sc.next();
        ArrayList<Integer> starts = new ArrayList<Integer>();
        ArrayList<Character> chars = new ArrayList<Character>();
        chars.add(s.charAt(0));
        starts.add(0);
        for (int i = 1; i < s.length(); i++) {
            if (!chars.contains(s.charAt(i))) {
                if (starts.size() < k) {
                    starts.add(i);
                    chars.add(s.charAt(i));
                }
            }
        }
        if (k > starts.size()) {
            System.out.println("NO");
            return;
        }
        starts.add(s.length());
        System.out.println("YES");
        for (int i = 1; i < starts.size(); i++) {
            System.out.println(s.substring(starts.get(i - 1), starts.get(i)));
        }
    }
}
