<p>You are given an array of non-negative integers <code>nums</code> and an integer <code>k</code>. In one operation, you may choose <strong>any</strong> element from <code>nums</code> and <strong>increment</strong> it by <code>1</code>.</p>

<p>Return<em> the <strong>maximum</strong> <strong>product</strong> of </em><code>nums</code><em> after <strong>at most</strong> </em><code>k</code><em> operations. </em>Since the answer may be very large, return it <b>modulo</b> <code>10<sup>9</sup> + 7</code>. Note that you should maximize the product before taking the modulo.&nbsp;</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [0,4], k = 5
<strong>Output:</strong> 20
<strong>Explanation:</strong> 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.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [6,3,3,2], k = 2
<strong>Output:</strong> 216
<strong>Explanation:</strong> 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.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length, k &lt;= 10<sup>5</sup></code></li>
	<li><code>0 &lt;= nums[i] &lt;= 10<sup>6</sup></code></li>
</ul>

class Solution:
    def maximumProduct(self, nums: List[int], k: int) -> int:
        M = 10 ** 9 + 7
        nums.sort()
        nums.append(math.inf)
        n = len(nums)
        prev = 0
        cnt = 0
        li = []
        for i in range(n):
            diff = nums[i] - prev
            tdiff = diff * cnt
            if tdiff <= k:
                k -= tdiff
                prev = nums[i]
                cnt += 1
            else:
                prev += k // cnt
                m = k % cnt
                li += [prev] * (cnt-m)
                li += [prev+1] * (m)
                li += nums[i:-1]
                break
        ret = 1
        for i in li:
            ret = ret * i % M
        return ret

        # M = 10 ** 9 + 7
        # heapq.heapify(nums)
        # for i in range(k):
        #     v = heapq.heappop(nums)
        #     heapq.heappush(nums, v+1)
        # ret = 1
        # for i in nums:
        #     ret = ret * i % M
        # return ret

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