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.*;

public class Main {

    /**
     * @param args the command line arguments
     */
    public static void main(String[] args) {
        int k,i,j,cs = 0;
        String inp;
        int[] cal = new int[110];
        int[] cnt = new int[50];
        
        Scanner sc = new Scanner(System.in);
        
        k = sc.nextInt();
        inp = sc.next();
        
        for(i = 0; i < 26; i++)
        {
            cnt[i] = 0;
        }
        
        for(i = 0; i < inp.length(); i++)
        {
            if(cnt[inp.charAt(i) - 'a'] == 0)
            {
                cal[i] = 1;
                cs++;
                cnt[inp.charAt(i) - 'a'] = 1;
            } else {
                cal[i] = 0;
            }
            
            //System.out.println("cal[" + i + "] = " + cal[i]);
        }
        
        if(cs >= k)
        {
            System.out.println("YES");
            
            i = 0;
            
            for(j = 1; j < k; j++)
            {
                System.out.print(inp.charAt(i++));
                
                for(; cal[i] == 0; i++)
                {
                    System.out.print(inp.charAt(i));
                }
                
                System.out.print("\n");
            }
            
            for(; i < inp.length(); i++)
            {
                System.out.print(inp.charAt(i));
            }
        } else {
            System.out.println("NO");
        }
    }
    
}
