<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 collections import deque
from typing import List
from math import pow

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        n = len(nums)
        maxNum = max(nums)
        minPrimeFactors = self.sieveEratosthenes(maxNum + 1)
        primeScores = self.getPrimeScores(nums, minPrimeFactors)
        ans = 1
        left = [-1] * n
        right = [n] * n
        stack = deque()

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

        stack.clear()

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

        numAndIndexes = [(num, i) for i, num in enumerate(nums)]
        numAndIndexes.sort(key=lambda x: (-x[0], x[1]))


        for num, i in numAndIndexes:
            rangeCount = (i - left[i]) * (right[i] - i)
            actualCount = min(rangeCount, k)
            k -= actualCount
            ans = (ans * self.modPow(num, actualCount)) % self.kMod

        return ans

    kMod = 1000000007

    def modPow(self, x, n):
        if n == 0:
            return 1
        if n % 2 == 1:
            return x * self.modPow(x, n - 1) % self.kMod
        return self.modPow(x * x % self.kMod, n // 2)

    def sieveEratosthenes(self, n):
        minPrimeFactors = [0] * (n + 1)
        for i in range(2, n + 1):
            minPrimeFactors[i] = i
        for i in range(2, int(n ** 0.5) + 1):
            if minPrimeFactors[i] == i:
                for j in range(i * i, n + 1, i):
                    minPrimeFactors[j] = min(minPrimeFactors[j], i)
        return minPrimeFactors

    def getPrimeScores(self, nums, minPrimeFactors):
        primeScores = [0] * len(nums)
        for i, num in enumerate(nums):
            primeScores[i] = self.getPrimeScore(num, minPrimeFactors)
        return primeScores

    def getPrimeScore(self, num, minPrimeFactors):
        primeFactors = set()
        while num > 1:
            divisor = minPrimeFactors[num]
            primeFactors.add(divisor)
            while num % divisor == 0:
                num //= divisor
        return len(primeFactors)

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