<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(object):
    def maximumScore(self, nums, k):
        """
        :type nums: List[int]
        :type k: int
        :rtype: int
        """
        MOD = 10 ** 9 + 7
        n = len(nums)
        factor=[0]*n
        scores=[]
        for i in range(n):
            score=self.prime_factors_count(nums[i])
            factor[i]=score
            #scores.append([score, nums[i], i])


        
        #scores.sort(key=lambda x: (-x[1]))

        result = 1
        stack = []
        left_bounds = [-1] * n
        for idx in range(n):
            while stack and factor[stack[-1]] < factor[idx]:
                stack.pop()
            left_bounds[idx] = stack[-1] if stack else -1
            stack.append(idx)

        # 使用单调栈找到每个因子的右边界
        stack = []
        right_bounds = [n] * n
        for idx in range(n-1, -1, -1):
            while stack and factor[stack[-1]] <= factor[idx]:
                stack.pop()
            right_bounds[idx] = stack[-1] if stack else n
            stack.append(idx)

        # 使用左右边界计算每个元素的times
        times = [(right_bounds[i] - i) * (i - left_bounds[i]) for i in range(n)]
        pairs = sorted(zip(nums, times), key=lambda x: -x[0])

        result = 1
        for num, time in pairs:
            use_times = min(k, time)
            result = (result * self.quick_pow(num, use_times, MOD)) % MOD
            k -= use_times
            if k == 0:
                return result 
                break 

            
            
    def prime_factors_count(self,n):
        j, res = 2, []
        while n > 1:
            for i in range(j, int(sqrt(n + 0.05)) + 1):
                if n % i == 0:
                    n //= i
                    j = i
                    res.append(i)
                    break
            else:
                if n > 1:
                    res.append(n)
                    break

        return len(set(res))
    def quick_pow(self,a, b, MOD):
        result = 1
        while b:
            if b & 1:
                result = (result * a) % MOD
            a = (a * a) % MOD
            b >>= 1
        return result
        

39300