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.Scanner;
import java.io.OutputStream;
import java.io.IOException;
import java.io.PrintWriter;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 * @author pankajmore
 */
public class Main {
	public static void main(String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		Scanner in = new Scanner(inputStream);
		PrintWriter out = new PrintWriter(outputStream);
		TaskA solver = new TaskA();
		solver.solve(1, in, out);
		out.close();
	}
}

class TaskA {
    public void solve(int testNumber, Scanner in, PrintWriter out) {
        int K = in.nextInt();
        String s = in.next();
        char[] ch = s.toCharArray();
        int N = s.length();
        int A[] = new int[26];
        int d = 0;
        for (char c : ch) {
            if(A[c-97] == 0){
                A[c-97] = 1;
                d += 1;
            }
        }
        if(d < K){
            out.println("NO");
            return;
        }
        else{
            out.print("YES");
            int i = 0,flag = 0;
            int B[] = new int[26];
            for(char c : ch){
                if(i==K){
                    out.print(c);
                    continue;
                }
                if(B[c-97] == 0){
                    B[c-97] = 1;
                    out.println();
                    i++;
                    out.print(c);
                }
                else{
                    out.print(c);
                }
               
            }
        }

    }
}

