<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 ans = 0;
    vector<int> memo;
    void DFS(vector<int>& is_visited, vector<vector<int>>& gcd_table){
        int now = -1;
        for(int i=0; i<is_visited.size(); i++){
            if(!is_visited[i]){
                now = i;
                break;
            }
        }
        if(now == -1){
            int M = 0;
            vector<int> tmp = memo;
            sort(tmp.begin(), tmp.end());
            for(int i=0; i<tmp.size(); i++){
                M += tmp[i] * (i+1);
            }
            ans = max(ans, M);
            return;
        }
        is_visited[now] = 1;
        for(int i=0; i<is_visited.size(); i++){
            if(!is_visited[i]){
                is_visited[i] = 1;
                memo.push_back(gcd_table[now][i]);
                DFS(is_visited, gcd_table);
                memo.pop_back();
                is_visited[i] = 0;
            }
        }
        is_visited[now] = 0;
    }
    int maxScore(vector<int>& nums) {
        ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);
        vector<vector<int>> gcd_table(nums.size(), vector<int>(nums.size()));
        vector<int> is_visited(nums.size(), 0);
        for(int i=0; i<nums.size(); i++){
            for(int j=i+1; j<nums.size(); j++){
                gcd_table[i][j] = gcd(nums[i], nums[j]);
                gcd_table[j][i] = gcd(nums[i], nums[j]);
            }
        }
        DFS(is_visited, gcd_table);
        return ans;
    }
};
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