<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>


n = int(1e5+10)
stk = [0] * (n)

prime = [True] * (n)
for i in range(2, n):
    if prime[i]:
        for j in range(i, n, i):
            prime[j]= False
            stk[j] += 1


class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        n = len(nums)
        MOD = int(1e9+7)
        arr = []
        arr2 = []
        for i, num in enumerate(nums):
            arr.append( stk[num] )
            arr2.append( (num, i) )

        left = []
        temp = []

        for i,num in enumerate(arr):
            while temp and arr[temp[-1]] < num:
                temp.pop()
            left.append(temp[-1] if temp else -1)
            temp.append(i)

        right = [None] * n
        temp = []
        for i in range(n-1, -1, -1):
            num = arr[i]
            while temp and arr[temp[-1]] <= num:
                temp.pop()
            right[i] = temp[-1] if temp else n
            temp.append(i)



        arr2.sort(reverse=True)
        arr3 = []

        for i in range(n):
            num = arr2[i][0]
            i2 = arr2[i][1]
            i1,i3 = left[i2],right[i2]
            arr3.append( (num, (i2-i1)*(i3-i2)) )

        ans = 1
        arr3.sort(reverse=True)
        for i in range(n):
            if arr3[i][1] < k:
                ans *= self.myPow(arr3[i][0], arr3[i][1]) % MOD
                k -= arr3[i][1]
            else:
                ans *= self.myPow(arr3[i][0], k) % MOD
                return ans % MOD

    def myPow(self, x: float, n: int) -> float:
        MOD = int(1e9 + 7)
        def quickMul(N):
            if N == 0:
                return 1
            y = quickMul(N // 2) % MOD
            return y * y if N % 2 == 0 else y * y * x

        return quickMul(n) % MOD  if n >= 0 else 1.0 / quickMul(-n)
46400