You are given an array of non-negative integers numsand an integer k. In one operation, you may choose any element from numsand increment it by 1.

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Return the maximum product of numsafter at most koperations. Since the answer may be very large, return it modulo 109 + 7. Note that you should maximize the product before taking the modulo. 

Example 1:

Input: nums = [0,4], k = 5

Output: 20

Explanation: Increment the first number 5 times.Now nums = [5, 4], with a product of 5 * 4 = 20.It can be shown that 20 is maximum product possible, so we return 20.Note that there may be other ways to increment nums to have the maximum product.

Example 2:

Input: nums = [6,3,3,2], k = 2

Output: 216

Explanation: Increment the second number 1 time and increment the fourth number 1 time.Now nums = [6, 4, 3, 3], with a product of 6 * 4 * 3 * 3 = 216.It can be shown that 216 is maximum product possible, so we return 216.Note that there may be other ways to increment nums to have the maximum product.

Constraints:

1 <= nums.length, k <= 105

0 <= nums[i] <= 106

如何让乘积最大，这里面数组里面的数越平均，得到的值就越大，所以我们每次加1的时候，加在最小的那个数上面就可以，所以我们就用优先队列来解决这个问题就可以了

下面是代码：

Runtime: 456 ms, faster than 12.87% of Java online submissions for Maximum Product After K Increments.

Memory Usage: 53.3 MB, less than 58.48% of Java online submissions for Maximum Product After K Increments.