<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>

class Solution:
  def maximumScore(self, nums: list[int], k: int) -> int:
    kMod = 1_000_000_007
    n = len(nums)
    ans = 1
    minPrimeFactors = self._sieveEratosthenes(max(nums) + 1)
    primeScores = [self._getPrimeScore(num, minPrimeFactors) for num in nums]
    # left[i] := the next index on the left (if any)
    #            s.t. primeScores[left[i]] >= primeScores[i]
    left = [-1] * n
    # right[i] := the next index on the right (if any)
    #             s.t. primeScores[right[i]] > primeScores[i]
    right = [n] * n
    stack = []

    # Find the next indices on the left where `primeScores` are greater or equal.
    for i in reversed(range(n)):
      while stack and primeScores[stack[-1]] <= primeScores[i]:
        left[stack.pop()] = i
      stack.append(i)

    stack = []

    # Find the next indices on the right where `primeScores` are greater.
    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)]

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

    for num, i in sorted(numAndIndexes, key=lambda x: (-x[0], x[1])):
      # nums[i] is the maximum value in the range [left[i] + 1, right[i] - 1]
      # So, there are (i - left[i]) * (right[i] - 1) ranges where nums[i] will
      # be chosen.
      rangeCount = (i - left[i]) * (right[i] - i)
      actualCount = min(rangeCount, k)
      k -= actualCount
      ans *= modPow(num, actualCount)
      ans %= kMod

    return ans

  def _sieveEratosthenes(self, n: int) -> list[int]:
    """Gets the minimum prime factor of i, where 2 <= i <= n."""
    minPrimeFactors = [i for i in range(n + 1)]
    for i in range(2, int(n**0.5) + 1):
      if minPrimeFactors[i] == i:  # `i` is prime.
        for j in range(i * i, n, i):
          minPrimeFactors[j] = min(minPrimeFactors[j], i)
    return minPrimeFactors

  def _getPrimeScore(self, num: int, minPrimeFactors: list[int]) -> int:
    primeFactors = set()
    while num > 1:
      divisor = minPrimeFactors[num]
      primeFactors.add(divisor)
      while num % divisor == 0:
        num //= divisor
    return len(primeFactors)
49300