<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>

def get_prime(n): # 埃拉特斯特尼筛,时间复杂度为 O(nloglogn)
    prime = [True] * n
    prime[0] = prime[1] = True # not necessarily needed if we are only interested in the prime numbers which are kept in ans[]
    ans = []

    for i in range(2, n):
        if prime[i]:
            ans.append(i)
            #print(i, i*i, n)
            for j in range(i*i, n, i):
                prime[j] = False
    return ans

prime = get_prime(10**5+2)
prime_set = set(prime)

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        
        MOD = 10 ** 9 + 7
        
        def count_distinct_prime_factors(n):
            if n in prime_set:
                return 1
            
            distinct_prime_factors = set()
            divisor = 0
            
            while n > 1:
                while n % prime[divisor] == 0:
                    distinct_prime_factors.add(prime[divisor])
                    n //= prime[divisor]
                divisor += 1

            return len(distinct_prime_factors)
        
        left, right = [], [] 
        # left: how many numbers on the left of the current index that has a smaller prime score
        # right: how many numbers on the right of the current index that has a smaller or equal prime score
        
        prime_score = []
        
        for n in nums:
            # get all the prime factors
            prime_score.append(count_distinct_prime_factors(n))
        
        stack = []
        for i, n in enumerate(prime_score):
            while stack and prime_score[stack[-1]] < n:
                stack.pop()
            dist = i - stack[-1] if stack else i + 1
            left.append(dist)
            stack.append(i)
        
        stack = []
        for i in reversed(range(len(prime_score))):
            n = prime_score[i]
            while stack and prime_score[stack[-1]] <= n:
                stack.pop()
            dist = stack[-1] - i if stack else len(prime_score) - i
            right.append(dist)
            stack.append(i)
        
        right = right[::-1]
        
        #print(prime_score)
        #print('left', left)
        #print('right', right)
        
        values = [(n, i) for i, n in enumerate(nums)]
        values.sort()
        #print('values', values)
        
        res = 1
        tot = k
        while values and tot:
            n, i = values.pop()
            l, r = left[i], right[i]
            cnt = min(l * r, tot)
            res *= pow(n, cnt, MOD)
            res %= MOD
            tot -= cnt
        
        return res % MOD
34900