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;

/**
 * Created by poly on 4/13/15.
 */
public class Main {
    public static void main(String argv[]) {
        Scanner scanner = new Scanner(System.in);
        int n = scanner.nextInt();
        String q = scanner.next();
        List<String> ans = new ArrayList<String>();
        boolean[] used = new boolean[26];
        int start = -1;
        for (int i = 0; i < q.length(); ++i) {
            int c = q.charAt(i) - 'a';
            if (!used[c]) {
                if (start != -1) {
                    ans.add(q.substring(start, i));
                }
                if (ans.size() == n - 1) {
                    ans.add(q.substring(i));
                    break;
                }
                start = i;
                used[c] = true;
            }
        }
        if (ans.size() == n) {
            System.out.println("YES");
            for (String s : ans) System.out.println(s);
        } else {
            System.out.println("NO");
        }
    }
}
