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Intuition:\n\nThe problem seems to involve determining the products of elements within specific ranges of numbers, under certain modular arithmetic constraints.
# Intuition:\n\nThe problem seems to involve determining the products of elements within specific ranges of numbers, under certain modular arithmetic constraints. My initial thought is to decompose the problem into smaller parts: counting occurrences of specific bits and then using those counts to compute the results for each query efficiently.\n\n# Approach:\n\n1. **Helper Functions**:\n    - `cnt1(num)`: This function calculates the cumulative count of `1` bits up to a given number `num`.\n    - `acc0(num)`: This function calculates the cumulative count of `0` bits, weighted by their position, up to a given number `num`.\n\n2. **Main Counting Function**:\n    - `count(bound)`: This function determines the effective count of bits up to a boundary `bound`. It does this by:\n        - Finding the target number where the boundary is located.\n        - Calculating the cumulative count of bits up to that target.\n        - Adjusting the count based on the remaining bits within the boundary.\n\n3. **Processing Queries**:\n    - For each query, which consists of a range (`low`, `high`) and a modulus (`mod`), we:\n        - Use the `count` function to find the effective counts of bits within the range.\n        - Compute the product of `2` raised to the power of the difference in bit counts, modulo `mod`.\n\n4. **Return the Results**:\n    - Collect results for all queries in a list and return.\n\n# Complexity\n- **Time complexity**:\n  - The `cnt1` and `acc0` functions run in $$O(log n)$$ time since they iterate through the bit lengths of the numbers.\n  - The `count` function involves a binary search and bit-level operations, leading to a complexity of $$O(log^2 n)$$.\n  - Each query is processed in $$O(log^2 n)$$ time, making the overall complexity for `q` queries $$O(q log^2 n)$$.\n\n- **Space complexity**:\n  - The space complexity is $$O(1)$$ since we only use a constant amount of additional space for the computations.\n\n# Don\'t Forget to Upvote!!\uD83D\uDE3A\n\n# Python Code:\n\n```python\nclass Solution:\n    def findProductsOfElements(self, queries: List[List[int]]) -> List[int]:\n        \n        # Helper function to count the cumulative number of \'1\' bits up to num\n        def cnt1(num: int) -> int:\n            res = 0\n            for i in range(num.bit_length()):\n                cur = num % (1 << (i + 1))\n                res += (num - cur) // 2 \n                if cur >= 1 << i:\n                    res += cur + 1 - (1 << i)\n            return res\n        \n        # Helper function to count the cumulative weighted number of \'0\' bits up to num\n        def acc0(num: int) -> int:\n            res = 0\n            for i in range(num.bit_length()):\n                cur = num % (1 << (i + 1))\n                res += (num - cur) // 2 * i\n                if cur >= 1 << i:\n                    res += (cur + 1 - (1 << i)) * i\n            return res\n        \n        # Function to calculate the bit count for a given boundary\n        def count(bound: int) -> int:\n            # Find the target number where bound is located\n            target = bisect_left(range(bound), bound, key=cnt1)\n            rest = bound - cnt1(target - 1)\n            cnt = acc0(target - 1)\n            \n            # Adjust the count based on remaining bits\n            for i in range(target.bit_length()):\n                if target & (1 << i):\n                    cnt += i\n                    rest -= 1\n                    if not rest:\n                        break\n            return cnt\n        \n        # Process each query and compute the product of elements in the given range\n        return [pow(2, count(high + 1) - count(low), mod) for low, high, mod in queries]\n```\n
['Bit Manipulation', 'Python3']