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.
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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 <= 60给你一个正整数 p 。你有一个下标从 1 开始的数组 nums ，这个数组包含范围 [1, 2p - 1] 内所有整数的二进制形式（两端都 包含）。你可以进行以下操作 任意 次：
nums 中选择两个元素 x 和 y 。x 中的一位与 y 对应位置的位交换。对应位置指的是两个整数 相同位置 的二进制位。比方说，如果 x = 1101 且 y = 0011 ，交换右边数起第 2 位后，我们得到 x = 1111 和 y = 0001 。
请你算出进行以上操作 任意次 以后，nums 能得到的 最小非零 乘积。将乘积对 109 + 7 取余 后返回。
注意：答案应为取余 之前 的最小值。
\n\n\n\n
示例 1：
\n\n\n输入：p = 1\n输出：1\n解释：nums = [1] 。\n只有一个元素，所以乘积为该元素。\n\n\n
示例 2：
\n\n\n输入：p = 2\n输出：6\n解释：nums = [01, 10, 11] 。\n所有交换要么使乘积变为 0 ，要么乘积与初始乘积相同。\n所以，数组乘积 1 * 2 * 3 = 6 已经是最小值。\n\n\n
示例 3：
\n\n\n输入：p = 3\n输出：1512\n解释：nums = [001, 010, 011, 100, 101, 110, 111]\n- 第一次操作中，我们交换第二个和第五个元素最左边的数位。\n - 结果数组为 [001, 110, 011, 100, 001, 110, 111] 。\n- 第二次操作中，我们交换第三个和第四个元素中间的数位。\n - 结果数组为 [001, 110, 001, 110, 001, 110, 111] 。\n数组乘积 1 * 6 * 1 * 6 * 1 * 6 * 7 = 1512 是最小乘积。\n\n\n
\n\n
提示：
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