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 <cstdio>
#include <algorithm>
#include <cmath>
#include <vector>
#include <list>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cctype>
#include <cassert>
#include <climits>
#include <string>
#include <bitset>
#include <cfloat>
#include <random>
#include <iomanip>
#include <unordered_set>
using namespace std;
typedef long double ld;
typedef long long int ll;
typedef unsigned long long int ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<ll> vll;
typedef vector<pair<int,int> > vpii;
typedef vector<pair<ll,ll> > vpll;
typedef vector<vi> vvi;
typedef vector<vvi> vvvi;
typedef vector<vc> vvc;
typedef vector<vs> vvs;
typedef vector<vll> vvll;
typedef map<int,int> mii;
typedef set<int> si;
#define rep(i,n) for(ll i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define arep(i,v) for(auto i : v)
#define irep(it, stl) for(auto it = stl.begin(); it != stl.end(); it++)
#define drep(i,n) for(int i = (n) - 1; i >= 0; --i)
#define fin(ans) cout << (ans) << '\n'
#define STLL(s) strtoll(s.c_str(), NULL, 10)
#define mp(p,q) make_pair(p, q)
#define pb(n) push_back(n)
#define all(a) a.begin(), a.end()
#define rall(a) a.rbegin(), a.rend()
#define Sort(a) sort(a.begin(), a.end())
#define Rort(a) sort(a.rbegin(), a.rend())
#define MATHPI acos(-1)
#define fi first
#define se second
constexpr int dx[8] = {1, 0, -1, 0, 1, -1, -1, 1};
constexpr int dy[8] = {0, 1, 0, -1, 1, 1, -1, -1};
template <class T> inline bool chmax(T& a,T b){if(a<b){a=b;return 1;} return 0;}
template <class T> inline bool chmin(T& a,T b){if(a>b){a=b;return 1;} return 0;}
template <class T> inline void dump(T &v){irep(i, v){cout<<*i<<((i==--v.end())?'\n':' ');}}
inline string getline(){string s; getline(cin,s); return s;}
inline void yn(const bool b){b?fin("yes"):fin("no");}
inline void Yn(const bool b){b?fin("Yes"):fin("No");}
inline void YN(const bool b){b?fin("YES"):fin("NO");}
struct io{io(){ios::sync_with_stdio(false);cin.tie(0);}};
const int INF = INT_MAX;
constexpr ll LLINF = 1LL<<60;
constexpr ll MOD = 1000000007;
constexpr ld EPS = 1e-11;



int main() {
    
    int n;
    cin>>n;
    int k;
    cin>>k;
    ll bit=0;
    rep(i,k){
        int t;
        cin>>t;
        bit|=(1LL<<t);
    }
    rep(i,bit+1){
        if((i&bit)==i){
            cout<<i<<":";
            rep(j, 29) {
                if(i&(1LL<<j)){
                    cout<<" "<<j;
                }
            }
            cout<<endl;
        }
    }
}