<p>Given two integers <code>left</code> and <code>right</code>, return <em>the <strong>count</strong> of numbers in the <strong>inclusive</strong> range </em><code>[left, right]</code><em> having a <strong>prime number of set bits</strong> in their binary representation</em>.</p>

<p>Recall that the <strong>number of set bits</strong> an integer has is the number of <code>1</code>&#39;s present when written in binary.</p>

<ul>
	<li>For example, <code>21</code> written in binary is <code>10101</code>, which has <code>3</code> set bits.</li>
</ul>

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

<pre>
<strong>Input:</strong> left = 6, right = 10
<strong>Output:</strong> 4
<strong>Explanation:</strong>
6  -&gt; 110 (2 set bits, 2 is prime)
7  -&gt; 111 (3 set bits, 3 is prime)
8  -&gt; 1000 (1 set bit, 1 is not prime)
9  -&gt; 1001 (2 set bits, 2 is prime)
10 -&gt; 1010 (2 set bits, 2 is prime)
4 numbers have a prime number of set bits.
</pre>

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

<pre>
<strong>Input:</strong> left = 10, right = 15
<strong>Output:</strong> 5
<strong>Explanation:</strong>
10 -&gt; 1010 (2 set bits, 2 is prime)
11 -&gt; 1011 (3 set bits, 3 is prime)
12 -&gt; 1100 (2 set bits, 2 is prime)
13 -&gt; 1101 (3 set bits, 3 is prime)
14 -&gt; 1110 (3 set bits, 3 is prime)
15 -&gt; 1111 (4 set bits, 4 is not prime)
5 numbers have a prime number of set bits.
</pre>

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

<ul>
	<li><code>1 &lt;= left &lt;= right &lt;= 10<sup>6</sup></code></li>
	<li><code>0 &lt;= right - left &lt;= 10<sup>4</sup></code></li>
</ul>

vector<bool> Binary(int n) {
    //cout << n << " ";
    if (n == 0) return {false};
    vector<bool> ret;
    while (n > 0) {
        ret.push_back(n % 2);
        //cout << n % 2;
        n /= 2;
    }
    //cout << ": ";
    return ret;
}

int CountTrue(vector<bool> input) {
    int count = 0;
    for (bool i : input) {
        if (i) count++;
    }
    //cout << count << " ";
    return count;
}

bool IsPrime(int n) {
    if (n == 1 || n == 0) return false;
    for (int i = 2; i * i <= n; i++) {
        if (n % i == 0) return false;
    }
    return true;
}

class Solution {
public:
    int countPrimeSetBits(int left, int right) {
        int ret = 0;
        for (int i = left; i <=right; i++) {
            if (IsPrime(CountTrue(Binary(i)))) {
                ret++;
                //cout << " True." << endl;
            } else {
                //cout << " False." << endl;
            }
        }
        return ret;
    }
};
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