<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:
        """
        Q: given a single x, we would like to know in how many sub-array this could be the num chose
        
        Condition:
        1. we would like to have maximum num as possible -> starting from the first num => trying to find a O(1) query way
        2. the num must be with smallest index in the subarr -> possibly consider extend to left, right?
        
        如果不考虑任何条件 一个数可以在 (i+1)*(n-i-1)个数组里？
        
        所以往右，停止扫描的条件是遇到了更大的primescore
        往左，停止扫描的条件是遇到了相等或更大的primescore
        """
        
        def next_great_element_right(arr):
            n = len(arr)
            res = [n] * n
            # keep non-increasing monotonic array, like 4, 1, 3, 2
            stack = []
            for i in range(n):
                while stack and arr[i] > arr[stack[-1]]:
                    res[stack.pop()] = i
                stack.append(i)
            return res

        def next_great_element_left(arr):
            n = len(arr)
            res = [-1] * n
            stack = []
            for i in range(n - 1, -1, -1):
                while stack and arr[i] >= arr[stack[-1]]:
                    res[stack.pop()] = i
                stack.append(i)
            return res

        # Primes = {}
        def primelength(n):
            if n == 1:
                return 0
            # if n in Primes:
            #     return Primes[n]
            i = 2
            f = set()
            while i * i <= n:
                if n % i:
                    i += 1
                else:
                    n //= i
                    f.add(i)
            if n > 1:
                f.add(n)
            # Primes[n] = len(f)
            return len(f)
        
        P = []
        P0 = []
        for i, n in enumerate(nums):
            # print(primelength(n))
            x = primelength(n)
            P0.append(x)
            P.append((-n, -x, i))
            
        # P.sort(key = lambda x:(x[0], x[1], -x[2]), reverse=True)
        heapq.heapify(P)
        # print(P0)
        left = next_great_element_left(P0)
        right = next_great_element_right(P0)
        
        res = 1
        M = 10 ** 9 + 7
        while P:
            cur = heapq.heappop(P)
            n, score, idx = -cur[0], -cur[1], cur[2]
            # print(n, idx)
            l, r = left[idx], right[idx]
            # print(idx, nums[idx], score, l, nums[l], r)
            cl = (idx - (l + 1) + 1) 
            cr = (r - 1 - idx + 1)
            cur_sub = cl * cr
            x = min(cur_sub, k)
            res = res * pow(n, x, M) % M
            # res = res * (n ** x) % M
            k -= x
            if k <= 0:
                break
            
            # [l + 1, ... idx], [idx, ... r - 1]
            
        
        
        return res % M
41600