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 CPP):

Solution:
#include <iostream>
#include <string>
#include <algorithm>
#include <functional>
#include <vector>
#include <utility>
#include <cstring>
#include <iomanip>
#include <numeric>
#include <cmath>
#include <queue>
#include <map>
using namespace std;
typedef long long ll;
const int INF = 1<<30; 
const int MOD = 1e9 + 7;
const int dy[] = {1, 0, -1, 0};
const int dx[] = {0, 1, 0, -1};
int main()
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n, k;
    cin >> n >> k;
    vector<int> v;
    for(int i = 0; i < k; i++)
    {
        int x;
        cin >> x;
        v.push_back(x);
    }
    for(int bit = 0; bit < (1<<k); bit++)
    {
        int val = 0;
        vector<int> w;
        for(int i = 0; i < k; i++)
        {
            if(bit & (1<<i))
            {
                val |= (1<<v[i]);
                w.push_back(v[i]);
            }
        }
        cout << val << ":";
        for(int i = 0; i < w.size(); i++) cout << " " << w[i];
        cout << endl;
    }
    return 0;
}