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.PrintWriter;
import java.util.Scanner;
public class Codef {
    public static void main(String[] args) {
        Scanner in = new Scanner(System.in);
        try (PrintWriter out = new PrintWriter(System.out)) {
        int n = in.nextInt();
        String str = in.next();
        char[] ch = str.toCharArray();		
        int[] a = new int[26];
        int[] b = new int[n + 1];
        b[n] = str.length();
        int count = 0;
        for(int i = 0; i < str.length(); ++i){
            if( a[ch[i]-'a'] == 0){
                a[ch[i]-'a'] ++;
                b[count] = i;
                count++;
            }
            if(count == n)
                 break;
        }
        if( n <= count){
            out.println("YES");
            for(int i = 0; i < n; ++i){
                for(int j = b[i]; j < b[i + 1]; ++j){
                    out.print(ch[j]);
                }
               out.println();
            }			
        }
        else{
                out.println("NO");
        }
        }
       }
    }