// Use floor(log2(num)) + 1 to prevent overflow of multiplication.
    int productBits = LongMath.log2(product, FLOOR) + 1;
    int bits = LongMath.log2(startingNumber, FLOOR) + 1;
    // Check for the next power of two boundary, to save us a CLZ operation.
    int nextPowerOfTwo = 1 << (bits - 1);

    // Iteratively multiply the longs as big as they can go.
    for (long num = startingNumber; num <= n; num++) {