<p>Given an integer array <code>nums</code>, return <em>the sum of divisors of the integers in that array that have exactly four divisors</em>. If there is no such integer in the array, return <code>0</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [21,4,7]
<strong>Output:</strong> 32
<strong>Explanation:</strong> 
21 has 4 divisors: 1, 3, 7, 21
4 has 3 divisors: 1, 2, 4
7 has 2 divisors: 1, 7
The answer is the sum of divisors of 21 only.
</pre>

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

<pre>
<strong>Input:</strong> nums = [21,21]
<strong>Output:</strong> 64
</pre>

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

<pre>
<strong>Input:</strong> nums = [1,2,3,4,5]
<strong>Output:</strong> 0
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>4</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
</ul>

class Solution {
public:
    int sumFourDivisors(vector<int>& nums) {
        int ans = 0;

        auto go = [&](int x) -> int {
            vector<pair<int, int>> pf;
            for (int i = 2; i * i <= x; i++) {
                if (x % i != 0) continue;
                int pw = 0;
                while (x % i == 0) {
                    pw++, x /= i;
                }
                pf.emplace_back(i, pw);
            }
            if (x > 1) pf.emplace_back(x, 1);
            if (pf.size() > 2) return 0;
            if (pf.size() == 2) {
                if (pf[0].second > 1 || pf[1].second > 1) return 0;
                int p = pf[0].first, q = pf[1].first;
                return 1 + p + q + p * q;
            }
            if (pf[0].second != 3) return 0;
            int p = pf[0].first;
            return 1 + p + p * p + p * p * p;
        };

        for (auto n : nums) {
            if (n == 1) continue;
            ans += go(n);
        }
        return ans;
    }
};
22700