Problem: You are given a set $T$, which is a subset of $S$. The set $S$ consists of $0, 1, ... n-1$. Print all subsets of $T$. Note that we represent $0, 1, ... n-1$ as 00...0001, 00...0010, 00...0100, ..., 10...0000 in binary respectively and the integer representation of a subset is calculated by bitwise OR of existing elements.

Constraints

* $1 \leq n \leq 28$
* $0 \leq k \leq 18$
* $k \leq n$
* $0 \leq b_i < n$

Input

The input is given in the following format.


$n$
$k \; b_0 \; b_1 \; ... \; b_{k-1}$


$k$ is the number of elements in $T$, and $b_i$ represents elements in $T$.

Output

Print the subsets ordered by their decimal integers. Print a subset in the following format.


$d$: $e_0$ $e_1$ ...


Print ':' after the integer value $d$, then print elements $e_i$ in the subset in ascending order. Separate two adjacency elements by a space character.

Example

Input

4
2 0 2


Output

0:
1: 0
4: 2
5: 0 2

Solution (in JAVA):

Solution:
import java.util.Scanner;

public class Main  {

	public static void main(String[] args) {
		Scanner scan =new Scanner(System.in);

		int n = scan.nextInt();
		int k = scan.nextInt();
		
		int[] array = new int[k];
		for(int i=0; i<k; i++) {
			array[i] = scan.nextInt();
		}

		for(int i=0; i<Math.pow(2, k); i++){
			int[] array2 = new int[n];
			int tmp = i;
			int index = 0;
			for(int j=0; tmp>0; j++) {
				if(tmp %2 == 1) {
					array2[array[j]] = 1;
					index += Math.pow(2, array[j]);
				}
				tmp /= 2;
			}
			
			System.out.print(index + ":");
			for(int j=0; j<k; j++) {
				if(array2[array[j]] == 1) System.out.print(" " + array[j]);
			}
			System.out.println();
		}
	}
}