#include <iostream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <string>
#include <functional>
#include <bitset>
#include <cmath>

using namespace std;

int main()
{

    bitset<18> subset, T;
    
    int n, k, b;
    cin >> n >> k;

    while (k--)
    {
        cin >> b;
        T[b] = 1;
    }

    int upper = pow(2, n);
    for (int i = 0; i < upper; i++)
    {
        subset = i;
        if ((subset|T) == subset)
        {
            cout << i << ":";            
            for (int j = 0; j < n; j++)
            {
                if (subset[j])
                {
                    cout << ' ' <<  j;
                }
            }
            cout << endl;            
        }

    }
}

### Prompt

Create a solution in Cpp for the following problem:

You are given a set $T$, which is a subset of $U$. The set $U$ consists of $0, 1, ... n-1$. Print all sets, each of which is a subset of $U$ and includes $T$ as a subset. 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

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

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

5: 0 2
7: 0 1 2
13: 0 2 3
15: 0 1 2 3

### Response

```cpp
#include <iostream>
#include <iterator>
#include <vector>
#include <algorithm>
#include <string>
#include <functional>
#include <bitset>
#include <cmath>

using namespace std;

int main()
{

    bitset<18> subset, T;
    
    int n, k, b;
    cin >> n >> k;

    while (k--)
    {
        cin >> b;
        T[b] = 1;
    }

    int upper = pow(2, n);
    for (int i = 0; i < upper; i++)
    {
        subset = i;
        if ((subset|T) == subset)
        {
            cout << i << ":";            
            for (int j = 0; j < n; j++)
            {
                if (subset[j])
                {
                    cout << ' ' <<  j;
                }
            }
            cout << endl;            
        }

    }
}

```