You are given a positive integer p. Consider an array nums (1-indexed) that consists of the integers in the inclusive range [1, 2p - 1] in their binary representations. You are allowed to do the following operation any number of times:
x and y from nums.x and swap it with its corresponding bit in y. Corresponding bit refers to the bit that is in the same position in the other integer.For example, if x = 1101 and y = 0011, after swapping the 2nd bit from the right, we have x = 1111 and y = 0001.
Find the minimum non-zero product of nums after performing the above operation any number of times. Return this product modulo 109 + 7.
Note: The answer should be the minimum product before the modulo operation is done.
\n\n\n
Example 1:
\n\n\nInput: p = 1\nOutput: 1\nExplanation: nums = [1].\nThere is only one element, so the product equals that element.\n\n\n
Example 2:
\n\n\nInput: p = 2\nOutput: 6\nExplanation: nums = [01, 10, 11].\nAny swap would either make the product 0 or stay the same.\nThus, the array product of 1 * 2 * 3 = 6 is already minimized.\n\n\n
Example 3:
\n\n\nInput: p = 3\nOutput: 1512\nExplanation: nums = [001, 010, 011, 100, 101, 110, 111]\n- In the first operation we can swap the leftmost bit of the second and fifth elements.\n - The resulting array is [001, 110, 011, 100, 001, 110, 111].\n- In the second operation we can swap the middle bit of the third and fourth elements.\n - The resulting array is [001, 110, 001, 110, 001, 110, 111].\nThe array product is 1 * 6 * 1 * 6 * 1 * 6 * 7 = 1512, which is the minimum possible product.\n\n\n
\n
Constraints:
\n\n1 <= p <= 60Compiled with clang 11 using the latest C++ 17 standard.
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