Explained bit manipulation, clean code
Approach

For each query [L, R, MOD], we:

Calculate counts of powers of 2 up to L and R+1
Use difference array technique to get counts in range [L, R]
Compute
# Approach
1. For each query [L, R, MOD], we:
   - Calculate counts of powers of 2 up to L and R+1
   - Use difference array technique to get counts in range [L, R]
   - Compute product using modular exponentiation

2. The core bit manipulation magic happens in `calculate_pow2_counts_till_value`:
   - Visualize numbers as rows in binary matrix
   - For each column k (power of 2):
     ```
     Row Binary  k=0  k=1  k=2
      0  000     0    0    0
      1  001     1    0    0
      2  010     0    1    0
      3  011     1    1    0
      4  100     0    0    1
     ```
   - Observe that 1s appear in periods of size 2^(k+1)
   - Count complete periods with `(M >> (k + 1)) << k`
   - Handle incomplete period by checking k-th bit and remainder

# Complexity
- Time complexity: $$O(Q \cdot \log^2 M)$$ where Q is number of queries and M is maximum value
  - Each query requires binary search and bit counting operations
- Space complexity: $$O(\log M)$$ to store counts of powers of 2

# Code
```cpp []
class Solution {
public:
    vector<int> findProductsOfElements(vector<vector<long long>>& queries) {
        vector<int> res(queries.size());
        transform(queries.begin(), queries.end(), res.begin(),
                  [](const auto& q) { return calculate_product(q[0], q[1], q[2]); });
        return res;
    }

private:
    static int calculate_product(long long q_beg, long long q_end, long long q_mod) {
        vector<long long> pow_count_beg = calculate_pow2_counts(q_beg);
        vector<long long> pow_count_end = calculate_pow2_counts(q_end + 1);
        for (int i = 0; i < pow_count_beg.size(); ++i)
            pow_count_end[i] -= pow_count_beg[i];
        long long res = 1;
        for (int i = 1; i < pow_count_end.size(); ++i)
            res = res * qpow(1LL << i, pow_count_end[i], q_mod) % q_mod;
        return res % q_mod;
    }

    /** Calculates the count of each power of 2 in powerful arrays up to given size.
     * 1. Binary search to find largest complete powerful array that fits
     * 2. Calculate counts for complete arrays
     * 3. Add remaining powers from the last incomplete array */
    static vector<long long> calculate_pow2_counts(long long big_nums_prefix_size) {
        long long beg = 0, end = big_nums_prefix_size + 1;
        while (beg < end) {
            long long mid = (beg + end) / 2;
            if (calculate_pow2_counts_till_value(mid) > big_nums_prefix_size)
                end = mid;
            else
                beg = mid + 1;
        }
        vector<long long> pow2_counts;
        calculate_pow2_counts_till_value(beg - 1, &pow2_counts);

        // Add remaining value i.e. beg to complete size to big_nums_prefix_size
        long long size = accumulate(pow2_counts.begin(), pow2_counts.end(), 0LL);
        for (int i = 0; beg > 0 && size < big_nums_prefix_size; ++i, beg >>= 1) {
            if (beg & 1) {
                if (pow2_counts.size() < i + 1)
                    pow2_counts.resize(i + 1);
                ++pow2_counts[i];
                ++size;
            }
        }

        return pow2_counts;
    }

    /** Calculates the count of each power of 2 in powerful arrays up to a value.
     * NOTE: value is the encoded positive number i.e. value=5 encoded as fragment: [.., 1, 4, ..]
     * i.e. every value would be encoded into one or many elements; thus idx-es of value are > value
     * @param out Optional vector to store the counts
     * @return Total count of all powers of 2
     * Uses bit manipulation. */
    static long long calculate_pow2_counts_till_value(long long last_present_value, vector<long long>* out = nullptr) {
        long long M = last_present_value + 1;
        long long res = 0;
        /** Calculate count of set bits (1s) in k-th column up to row M-1 in binary matrix:
         *  Row Binary  k=0  k=1  k=2
         *   0  000     0    0    0
         *   1  001     1    0    0
         *   2  010     0    1    0
         *   3  011     1    1    0
         *   4  100     0    0    1
         *   5  101     1    0    1
         *   etc... */
        for (long long k = 0; 1LL << k < M; ++k) {
            // For k-th column, pattern of 1s and 0s has period of 2^(k+1)
            // Each period contains 2^k ones
            // ⌊M/(2^(k+1))⌋ gives number of complete periods
            // Multiply by 2^k to get total ones in complete periods
            long long count = (M >> (k + 1)) << k;

            // Now handle the incomplete period at the end:
            // Check if M has 1 in k-th position ((M >> k) & 1)
            if ((M >> k) & 1)
                // If yes, we need to count ones in remainder:
                // M & (2^k - 1) gives remainder after last complete period
                count += M & ((1LL << k) - 1);

            res += count;
            if (out)
                out->push_back(count);
        }
        return res;
    }

    static long long qpow(long long b, long long p, long long mod) {
        b %= mod;
        long long res = 1;
        for (; p > 0; p >>= 1, b = b * b % mod)
            if (p & 1)
                res = res * b % mod;
        return res;
    }
};

```
['C++']