<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:
        unique = sorted(set(nums))
        count = {}
        for i in nums:
            count[i] = count.get(i,0)+1

        key = unique[0]
        val = count[unique[0]]

        if len(unique) == 1:
                key += k//val
                rem = k%val
                base = (key**(val-rem))*((key+1)**rem)
                return base%(10**9+7)

        i = 1
        while i<len(unique):
            n = unique[i]
            t = count[unique[i]]
            if (n-key)*val > k:
                key += k//val
                rem = k%val
                base = ((key**(val-rem))*((key+1)**rem)*(n**t))%(10**9+7)

                for j in range(i+1,len(unique)):
                    key = unique[j]
                    val = count[unique[j]]
                    base=(base*((key**val%(10**9+7))))%(10**9+7)
                return base
            else: 
                k-=(n-key)*val
                key = n
                val += t
                     
            i+=1

        key += k//val
        rem = k%val
        return ((key%(10**9+7))**(val-rem))*((key+1)%(10**9+7))**rem


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