<p>You are given an array <code>nums</code> of <code>n</code> positive integers and an integer <code>k</code>.</p>

<p>Initially, you start with a score of <code>1</code>. You have to maximize your score by applying the following operation at most <code>k</code> times:</p>

<ul>
	<li>Choose any <strong>non-empty</strong> subarray <code>nums[l, ..., r]</code> that you haven&#39;t chosen previously.</li>
	<li>Choose an element <code>x</code> of <code>nums[l, ..., r]</code> with the highest <strong>prime score</strong>. If multiple such elements exist, choose the one with the smallest index.</li>
	<li>Multiply your score by <code>x</code>.</li>
</ul>

<p>Here, <code>nums[l, ..., r]</code> denotes the subarray of <code>nums</code> starting at index <code>l</code> and ending at the index <code>r</code>, both ends being inclusive.</p>

<p>The <strong>prime score</strong> of an integer <code>x</code> is equal to the number of distinct prime factors of <code>x</code>. For example, the prime score of <code>300</code> is <code>3</code> since <code>300 = 2 * 2 * 3 * 5 * 5</code>.</p>

<p>Return <em>the <strong>maximum possible score</strong> after applying at most </em><code>k</code><em> operations</em>.</p>

<p>Since the answer may be large, return it modulo <code>10<sup>9 </sup>+ 7</code>.</p>

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

<pre>
<strong>Input:</strong> nums = [8,3,9,3,8], k = 2
<strong>Output:</strong> 81
<strong>Explanation:</strong> To get a score of 81, we can apply the following operations:
- Choose subarray nums[2, ..., 2]. nums[2] is the only element in this subarray. Hence, we multiply the score by nums[2]. The score becomes 1 * 9 = 9.
- Choose subarray nums[2, ..., 3]. Both nums[2] and nums[3] have a prime score of 1, but nums[2] has the smaller index. Hence, we multiply the score by nums[2]. The score becomes 9 * 9 = 81.
It can be proven that 81 is the highest score one can obtain.</pre>

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

<pre>
<strong>Input:</strong> nums = [19,12,14,6,10,18], k = 3
<strong>Output:</strong> 4788
<strong>Explanation:</strong> To get a score of 4788, we can apply the following operations: 
- Choose subarray nums[0, ..., 0]. nums[0] is the only element in this subarray. Hence, we multiply the score by nums[0]. The score becomes 1 * 19 = 19.
- Choose subarray nums[5, ..., 5]. nums[5] is the only element in this subarray. Hence, we multiply the score by nums[5]. The score becomes 19 * 18 = 342.
- Choose subarray nums[2, ..., 3]. Both nums[2] and nums[3] have a prime score of 2, but nums[2] has the smaller index. Hence, we multipy the score by nums[2]. The score becomes 342 * 14 = 4788.
It can be proven that 4788 is the highest score one can obtain.
</pre>

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

<ul>
	<li><code>1 &lt;= nums.length == n &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= k &lt;= min(n * (n + 1) / 2, 10<sup>9</sup>)</code></li>
</ul>

from typing import List
from collections import defaultdict
import heapq

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        MOD = 10**9 + 7

        def prime_factors_count(x):
            count = 0
            factor = 2
            while factor * factor <= x:
                if x % factor == 0:
                    count += 1
                    while x % factor == 0:
                        x //= factor
                factor += 1
            if x > 1:
                count += 1
            return count

        n = len(nums)
        prime_scores = [prime_factors_count(num) for num in nums]

        left = [-1] * n
        right = [n] * n
        stack = []

        for i in range(n):
            while stack and prime_scores[stack[-1]] < prime_scores[i]:
                stack.pop()
            if stack:
                left[i] = stack[-1]
            stack.append(i)

        stack = []
        for i in range(n - 1, -1, -1):
            while stack and prime_scores[stack[-1]] <= prime_scores[i]:
                stack.pop()
            if stack:
                right[i] = stack[-1]
            stack.append(i)

        candidates = []
        for i in range(n):
            count = (i - left[i]) * (right[i] - i)
            candidates.append((-nums[i], count))

        heapq.heapify(candidates)
        score = 1

        while k > 0 and candidates:
            num, count = heapq.heappop(candidates)
            num = -num
            use = min(count, k)
            score = (score * pow(num, use, MOD)) % MOD
            k -= use

        return score
38100