<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 get_prime_cnt(self, num):
        if num==1:
            return 0
        elif num<=3:
            return 1
        ret = 0
        i = 2
        history = None
        while i*i <= num:
            if num%i:
                i += 1
            else:
                if i != history:
                    ret += 1
                    history = i
                num //= i
        if num != history:
            ret += 1
        return ret
    
    def pow2(self, num, t):
        if t==1:
            return num
        res = 1
        while t:
            if t % 2 == 1:
                res *= num % int(math.pow(10,9)+7)
            num *= num % (int(math.pow(10,9)+7))
            t //= 2
        return res


    def maximumScore(self, nums: List[int], k: int) -> int:
        mp = {}
        for num in nums:
            if num not in mp:
                mp[num] = self.get_prime_cnt(num)
        score = 1
        n = len(nums)
        # left to right
        left_stack = []
        l = 0
        for i in range(n):
            if not left_stack or mp[nums[i]] <= mp[left_stack[-1][0]]:
                l = i
            else:
                while left_stack and mp[nums[i]] > mp[left_stack[-1][0]]:
                    _, l = left_stack.pop()
            left_stack.append((nums[i], l))
            nums[i] = [nums[i],l]
        # right to left
        right_stack = []
        r = n-1
        for i in range(n-1, -1, -1):
            if not right_stack or mp[nums[i][0]] < mp[right_stack[-1][0]]:
                r = i
            else:
                while right_stack and mp[right_stack[-1][0]] <= mp[nums[i][0]]:
                    _, r= right_stack.pop()
            right_stack.append((nums[i][0], r))
            rl = i -nums[i][1]+1
            rr = r - i+1
            nums[i] = [nums[i][0], rl*rr]
        nums = sorted(nums, key=lambda x:-x[0])
        i = 0
        while(i<n):
            if nums[i][0] == 1:
                return score
            cnt = nums[i][1]
            t = min(cnt, k)
            score *= self.pow2(nums[i][0], t)
            score %= int(math.pow(10,9)+7)
            k -= cnt
            i += 1
            if k<=0:
                return score
        return score
45600