<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>

import heapq

class Solution:
    def maximumProduct(self, nums, k):
        # Step 1: Initialize a min heap (priority queue)
        heapq.heapify(nums)
        
        # Step 2: Perform k increment operations
        for _ in range(k):
            smallest = heapq.heappop(nums)
            heapq.heappush(nums, smallest + 1)
        
        # Step 3: Calculate the product of the modified array
        product = 1
        modulo = 10**9 + 7
        for num in nums:
            product = (product * num) % modulo
        
        return product

# Example usage:
solution = Solution()
nums = [0, 4]
k = 5
print(solution.maximumProduct(nums, k))  # Output: 20

nums = [6, 3, 3, 2]
k = 2
print(solution.maximumProduct(nums, k))  # Output: 216
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