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.List;
import java.util.Scanner;

public class A544 {

    public static void main(String[] args) {
        Scanner input = new Scanner(System.in);
        int K = input.nextInt();
        char[] c = input.next().toCharArray();
        List<String> solution = new ArrayList<String>();
        StringBuilder next = new StringBuilder();
        boolean[] used = new boolean[26];
        for (int i=0; i<c.length; i++) {
            if (K > 1 && !used[c[i]-'a']) {
                used[c[i]-'a'] = true;
                if (next.length() > 0) {
                    solution.add(next.toString());
                    K--;
                    next.setLength(0);
                }
            }
            next.append(c[i]);
        }
        solution.add(next.toString());
        if (K != 1) {
            System.out.println("NO");
        } else {
            System.out.println("YES");
            for (String part : solution) {
                System.out.println(part);
            }
        }
    }

}
