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.*;
import java.util.*;

public class Main{
    public static void main(String args[]) throws IOException{
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        int n = Integer.parseInt(br.readLine());
        String entrada = br.readLine().trim();
        char[] arr = entrada.toCharArray();
        int[] ind = new int[n];
        int currInd = -1;
        ind[0] = 0;
        currInd = 0;
        for(int i=0; i<arr.length && currInd != n-1; i++){
            boolean posible = true;
            for(int j=0; j<=currInd; j++){
                if(arr[i] == arr[ind[j]]){
                    posible = false;
                    break;
                }
            }
            if(posible){
                ind[++currInd] = i;
            }
        }
        if (currInd != n-1){
            System.out.println("NO");
        }else{
            System.out.println("YES");
            for(int i=0; i<n-1; i++){
                System.out.println(entrada.substring(ind[i],ind[i+1]));
            }
            System.out.println(entrada.substring(ind[currInd]));
        }
    }
}