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.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.HashSet;
import java.util.Set;

public class Main {
    public static void main(String[] args) throws IOException {

        BufferedReader bufferedReader = new BufferedReader(
                new InputStreamReader(System.in));
        int x = Integer.valueOf(bufferedReader.readLine());
        String string = bufferedReader.readLine();
        Set<Character> set = new HashSet<Character>();
        ArrayList<Integer> list = new ArrayList<Integer>();
        int index = 0;
        for (char c : string.toCharArray()) {
            if (!set.contains(c)) {
                set.add(c);
                list.add(index);
                index++;
            } else {
                index++;
            }
        }
        if (set.size() >= x) {
            System.out.println("YES");
            for (int i = 0, iteration = 1; i < list.size() - 1; i++, iteration++) {
                if (iteration == x) {
                    System.out.println(string.substring(list.get(i)));
                    return;
                }
                System.out.println(string.substring(list.get(i),
                        list.get(i + 1)));
            }
            System.out.println(string.substring(list.get(list.size() - 1),
                    string.length()));
        } else {
            System.out.println("NO");
        }

    }
}