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 <vector>
#include <algorithm>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <numeric>
#include <climits>
#include <iterator>
#include <iomanip>
#include <stack>
#include <set>
#include <cstdio>
#include <bitset>
#include <functional>
#include <tuple>
#include <list>
#include <deque>
using namespace std;
typedef pair<int, int> P;

#define FOR(i, a, n) for (ll i = (ll)a; i<(ll)n; ++i)
#define REP(i, n) FOR(i, 0, n)

typedef long long ll;
typedef vector<int> VI;

int main() {
    cin.tie(0);
   	ios::sync_with_stdio(false);
    
    int n; cin >> n;
    int bit=0;
    int k; cin >> k;
    for(int i=0; i<k; ++i){
        int b; cin >> b;
        bit|=(1<<b);
    }
    vector<int> elem;
    for(int sub = bit; ; sub = (sub-1)&bit) {
        elem.push_back(sub);
        if(!sub) break;
    }
    
    sort(elem.begin(), elem.end());
    
    REP(i, elem.size()){
        cout << elem[i] << ":";
        REP(j, n) if((elem[i]>>j)&1) cout << " " << j;
        cout << endl;
    }
	return 0;
}