You are given an array of non-negative integers nums and an integer k. In one operation, you may choose any element from nums and increment it by 1.
Return the maximum product of nums after at most k operations. Since the answer may be very large, return it modulo 109 + 7. Note that you should maximize the product before taking the modulo.
\n
Example 1:
\n\n\nInput: nums = [0,4], k = 5\nOutput: 20\nExplanation: Increment the first number 5 times.\nNow nums = [5, 4], with a product of 5 * 4 = 20.\nIt can be shown that 20 is maximum product possible, so we return 20.\nNote that there may be other ways to increment nums to have the maximum product.\n\n\n
Example 2:
\n\n\nInput: nums = [6,3,3,2], k = 2\nOutput: 216\nExplanation: Increment the second number 1 time and increment the fourth number 1 time.\nNow nums = [6, 4, 3, 3], with a product of 6 * 4 * 3 * 3 = 216.\nIt can be shown that 216 is maximum product possible, so we return 216.\nNote that there may be other ways to increment nums to have the maximum product.\n\n\n
\n
Constraints:
\n\n1 <= nums.length, k <= 1050 <= nums[i] <= 106给你一个非负整数数组 nums 和一个整数 k 。每次操作，你可以选择 nums 中 任一 元素并将它 增加 1 。
请你返回 至多 k 次操作后，能得到的 nums的 最大乘积 。由于答案可能很大，请你将答案对 109 + 7 取余后返回。
\n\n
示例 1：
\n\n输入：nums = [0,4], k = 5\n输出：20\n解释：将第一个数增加 5 次。\n得到 nums = [5, 4] ，乘积为 5 * 4 = 20 。\n可以证明 20 是能得到的最大乘积，所以我们返回 20 。\n存在其他增加 nums 的方法，也能得到最大乘积。\n\n\n
示例 2：
\n\n输入：nums = [6,3,3,2], k = 2\n输出：216\n解释：将第二个数增加 1 次，将第四个数增加 1 次。\n得到 nums = [6, 4, 3, 3] ，乘积为 6 * 4 * 3 * 3 = 216 。\n可以证明 216 是能得到的最大乘积，所以我们返回 216 。\n存在其他增加 nums 的方法，也能得到最大乘积。\n\n\n
\n\n
提示：
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