<p>You are given <code>nums</code>, an array of positive integers of size <code>2 * n</code>. You must perform <code>n</code> operations on this array.</p>

<p>In the <code>i<sup>th</sup></code> operation <strong>(1-indexed)</strong>, you will:</p>

<ul>
	<li>Choose two elements, <code>x</code> and <code>y</code>.</li>
	<li>Receive a score of <code>i * gcd(x, y)</code>.</li>
	<li>Remove <code>x</code> and <code>y</code> from <code>nums</code>.</li>
</ul>

<p>Return <em>the maximum score you can receive after performing </em><code>n</code><em> operations.</em></p>

<p>The function <code>gcd(x, y)</code> is the greatest common divisor of <code>x</code> and <code>y</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2]
<strong>Output:</strong> 1
<strong>Explanation:</strong>&nbsp;The optimal choice of operations is:
(1 * gcd(1, 2)) = 1
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [3,4,6,8]
<strong>Output:</strong> 11
<strong>Explanation:</strong>&nbsp;The optimal choice of operations is:
(1 * gcd(3, 6)) + (2 * gcd(4, 8)) = 3 + 8 = 11
</pre>

<p><strong class="example">Example 3:</strong></p>

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5,6]
<strong>Output:</strong> 14
<strong>Explanation:</strong>&nbsp;The optimal choice of operations is:
(1 * gcd(1, 5)) + (2 * gcd(2, 4)) + (3 * gcd(3, 6)) = 1 + 4 + 9 = 14
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= n &lt;= 7</code></li>
	<li><code>nums.length == 2 * n</code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution {
public:
    int check(vector<int> pair)
    {
        int sum = 0;
        int n = pair.size();
        sort(pair.rbegin(), pair.rend());

        for(auto p: pair)
        {
            sum = sum + p*n;
            n--;
        }
        return sum;
    }
    void sol(int i,vector<int>& nums, int& m,vector<int>& pair, int& res,vector<bool>& vs)
    {
        if(pair.size() == (m/2))
        {
            res = max(res,check(pair));
            return;
        }
        if(vs[i])
            sol(i+1,nums,m,pair,res,vs);
        else{    
            vs[i] = true;
            for(int j =i+1;j<m;j++)
            {
                if(!vs[j])
                {
                    vs[j] = true;
                    pair.push_back(__gcd(nums[i],nums[j]));
                    sol(i+1,nums,m,pair,res,vs);
                    vs[j] = false;
                    pair.pop_back();
                }
                
            }
            vs[i] = false;
        }
    }
    int maxScore(vector<int>& nums) {
        int m = nums.size();
        int n = m/2;
        vector<int> pair;
        int res = INT_MIN;
        vector<bool> vs(m, false);
        sol(0,nums,m,pair,res,vs);
        return res;
    }
};
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