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.*;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.List;
import java.util.Map;

public class Gots {

    public static long gcd(long a, long b) {
        return (b == 0) ? a : gcd(b, a % b);
    }

    public static void main(String[] args) throws IOException {
        BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
        PrintWriter writer = new PrintWriter(System.out);
        int k = Integer.parseInt(reader.readLine());
        //int[] arr = new int[n];
        String s = reader.readLine();
        Map<Character, Integer> map = new HashMap();
        List<Integer> indexes = new ArrayList();
        List<String> ans = new ArrayList();
        for (int i = 0; i < s.length(); i++) {
            if (!map.containsKey(s.charAt(i))) {
                map.put(s.charAt(i), i);
                indexes.add(i);
            }
        }
        if (indexes.size() < k) {
            writer.println("NO");
        } else {
            writer.println("YES");
            Collections.sort(indexes);
            for (int i = 0; i < k - 1; i++) {
                ans.add(s.substring(indexes.get(i), indexes.get(i + 1)));
            }
            ans.add(s.substring(indexes.get(k - 1), s.length()));
            for (int i = 0; i < ans.size(); i++) {
                writer.println(ans.get(i));
            }
        }

        writer.flush();

    }
}
