<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>

'''
nums = [6,3,3,2], k = 2
6 -> 18
3 -> 36
2 -> 54

6 -> 27
3 -> 54
'''
class Solution:
    def maximumProduct(self, nums: List[int], k: int) -> int:
        CONST = 10**9 + 7
        minheap = []
        for i in range(len(nums)):
            minheap.append((nums[i], i))
        
        heapq.heapify(minheap)
        
        while k > 0:
            val, index = heapq.heappop(minheap)
            heapq.heappush(minheap, (val+1, index))
            k -= 1

        product = 1
        while minheap:
            val, _ = heapq.heappop(minheap)
            product = (product * val) % CONST

        return product

        
1410