Problem:
Enumeration of Subsets III

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.

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.

Constraints

$1 \leq n \leq 28$

$0 \leq k \leq 18$

$k \leq n$

$0 \leq b_i < n$

Sample Input 1

4
2 0 2

Sample Output 1

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

Solution in JAVA (judge verdict: Accepted)

import java.util.Scanner;

public class Main {

	public static void main(String[] args) {
		// TODO 自動生成されたメソッド・スタブ
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		//int bitArray[] = new int[n];

		int k = sc.nextInt();
		int mask[] = new int[k];
		for(int i = 0; i < k; i++) {
			mask[i] = sc.nextInt();
		}

		for(int i = 0; i < Math.pow(2, k); i++) {  //maskがkの大きさなら2^kの大きさのビット列を作り、順列を求める
			int bitArray[] = new int[n];
			int temp = i;
			int index = 0;  //実際に表示されるindexはビットが1の値のjに対して2^mask[j]を足していったもの
			for(int j = 0; temp > 0; j++) {
				if(temp % 2 == 1) {
					bitArray[mask[j]] = 1;
					index += Math.pow(2, mask[j]);
				}
				temp /= 2;
			}

			//////////////////////////////////////////////////////////

			System.out.print(index + ":");  //順列のインデックス
			for(int j = 0; j < k; j++) {
				if(bitArray[mask[j]] == 1) System.out.print(" " + mask[j]);
			}
			System.out.println();
		}
	}

}