Find Products of Elements of Big Array
<p>The <strong>powerful array</strong> of a non-negative integer <code>x</code> is defined as the shortest sorted array of powers of two that sum up to <code>x</code>. The table below illustrates examples of how the <strong>powerful array</strong> is determined. It can be proven that the powerful array of <code>x</code> is unique.</p>

<table border="1">
	<tbody>
		<tr>
			<th>num</th>
			<th>Binary Representation</th>
			<th>powerful array</th>
		</tr>
		<tr>
			<td>1</td>
			<td>0000<u>1</u></td>
			<td>[1]</td>
		</tr>
		<tr>
			<td>8</td>
			<td>0<u>1</u>000</td>
			<td>[8]</td>
		</tr>
		<tr>
			<td>10</td>
			<td>0<u>1</u>0<u>1</u>0</td>
			<td>[2, 8]</td>
		</tr>
		<tr>
			<td>13</td>
			<td>0<u>11</u>0<u>1</u></td>
			<td>[1, 4, 8]</td>
		</tr>
		<tr>
			<td>23</td>
			<td><u>1</u>0<u>111</u></td>
			<td>[1, 2, 4, 16]</td>
		</tr>
	</tbody>
</table>

<p>The array <code>big_nums</code> is created by concatenating the <strong>powerful arrays</strong> for every positive integer <code>i</code> in ascending order: 1, 2, 3, and so on. Thus, <code>big_nums</code> begins as <code>[<u>1</u>, <u>2</u>, <u>1, 2</u>, <u>4</u>, <u>1, 4</u>, <u>2, 4</u>, <u>1, 2, 4</u>, <u>8</u>, ...]</code>.</p>

<p>You are given a 2D integer matrix <code>queries</code>, where for <code>queries[i] = [from<sub>i</sub>, to<sub>i</sub>, mod<sub>i</sub>]</code> you should calculate <code>(big_nums[from<sub>i</sub>] * big_nums[from<sub>i</sub> + 1] * ... * big_nums[to<sub>i</sub>]) % mod<sub>i</sub></code><!-- notionvc: a71131cc-7b52-4786-9a4b-660d6d864f89 -->.</p>

<p>Return an integer array <code>answer</code> such that <code>answer[i]</code> is the answer to the <code>i<sup>th</sup></code> query.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">queries = [[1,3,7]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[4]</span></p>

<p><strong>Explanation:</strong></p>

<p>There is one query.</p>

<p><code>big_nums[1..3] = [2,1,2]</code>. The product of them is 4. The result is <code>4 % 7 = 4.</code></p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">queries = [[2,5,3],[7,7,4]]</span></p>

<p><strong>Output:</strong> <span class="example-io">[2,2]</span></p>

<p><strong>Explanation:</strong></p>

<p>There are two queries.</p>

<p>First query: <code>big_nums[2..5] = [1,2,4,1]</code>. The product of them is 8. The result is <code>8 % 3 = 2</code>.</p>

<p>Second query: <code>big_nums[7] = 2</code>. The result is <code>2 % 4 = 2</code>.</p>
</div>

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

<ul>
	<li><code>1 &lt;= queries.length &lt;= 500</code></li>
	<li><code>queries[i].length == 3</code></li>
	<li><code>0 &lt;= queries[i][0] &lt;= queries[i][1] &lt;= 10<sup>15</sup></code></li>
	<li><code>1 &lt;= queries[i][2] &lt;= 10<sup>5</sup></code></li>
</ul>
Bit Manipulation; Array; Binary Search
m = 50
cnt = [0] * (m + 1)
s = [0] * (m + 1)
p = 1
for i in range(1, m + 1):
    cnt[i] = cnt[i - 1] * 2 + p
    s[i] = s[i - 1] * 2 + p * (i - 1)
    p *= 2


def num_idx_and_sum(x: int) -> tuple:
    idx = 0
    total_sum = 0
    while x:
        i = x.bit_length() - 1
        idx += cnt[i]
        total_sum += s[i]
        x -= 1 << i
        total_sum += (x + 1) * i
        idx += x + 1
    return (idx, total_sum)


def f(i: int) -> int:
    l, r = 0, 1 << m
    while l < r:
        mid = (l + r + 1) >> 1
        idx, _ = num_idx_and_sum(mid)
        if idx < i:
            l = mid
        else:
            r = mid - 1

    total_sum = 0
    idx, total_sum = num_idx_and_sum(l)
    i -= idx
    x = l + 1
    for _ in range(i):
        y = x & -x
        total_sum += y.bit_length() - 1
        x -= y
    return total_sum


class Solution:
    def findProductsOfElements(self, queries: List[List[int]]) -> List[int]:
        return [pow(2, f(right + 1) - f(left), mod) for left, right, mod in queries]
