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.Arrays;


public class Main {

    public static void main(String[]args) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String in = br.readLine();
        
        boolean[]v = new boolean[26];
        int n = Integer.parseInt(in);
        
        in = br.readLine();
        int c = 0;
        for(int i = 0;i<in.length();i++){
            if(!v[(int)in.charAt(i)-(int)'a']){
                v[(int)in.charAt(i)-(int)'a'] = true;
                c++;
            }
        }
        Arrays.fill(v, false);
        if(c>=n){
            StringBuilder b = new StringBuilder().append(in);
            StringBuilder ans = new StringBuilder();
            
            System.out.println("YES");
            c = 0;
            
            for(int i = 0;i<in.length();i++){
                if(!v[(int)in.charAt(i)-(int)'a'] && c<n){
                    v[(int)in.charAt(i)-(int)'a'] = true;
                    if(i>0)
                    ans.append('\n');

                        c++;
                    
                    ans.append(in.charAt(i));
                    
                }else{
                    ans.append(in.charAt(i));
                }
                
            }
            System.out.println(ans);
        }else{
            System.out.println("NO");
        }
        
        
    }
    
    
}
