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.*;
import java.io.*;
import static java.util.Arrays.*;
import static java.lang.Math.*;

public class Main {
    
    static final Scanner in = new Scanner(System.in);
    static final PrintWriter out = new PrintWriter(System.out,false);

    static void solve() {
        int k = in.nextInt();
        String s = in.next();
        int n = s.length();

        HashSet<Character> set = new HashSet<Character>();
        for (int i=0; i<n; i++) {
            set.add(s.charAt(i));
        }

        if (set.size() < k) {
            out.println("NO");
            return;
        }

        out.println("YES");
        if (n == 1) {
            out.println(s);
            return;
        }


        set.clear();
        set.add(s.charAt(0));
        int cnt = 0, ptr = 0;
        for (int i=1; i<n; i++) {
            if (!set.contains(s.charAt(i))) {
                if (++cnt == k) {
                    out.println(s.substring(ptr));
                    return;
                }
                set.add(s.charAt(i));
                out.println(s.substring(ptr,i));
                ptr = i;
            }
        }
        out.println(s.substring(ptr));
    }

    public static void main(String[] args) {
        long start = System.currentTimeMillis();

        solve();
        out.flush();

        long end = System.currentTimeMillis();
        //trace(end-start + "ms");
        in.close();
        out.close();
    }

    static void trace(Object... o) { System.out.println(deepToString(o));}
}