<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>

from sortedcontainers import SortedList
mod = int(1e9) + 7
N = int(1e5) + 1
IS_PRIME = [True] * N
IS_PRIME[0] = IS_PRIME[1] = IS_PRIME[2] = IS_PRIME[3] = True
IS_PRIME[0] = IS_PRIME[1] = False
for fac in range(2, N):
    if IS_PRIME[fac]:
        for val in range(fac * fac, N, fac):
            IS_PRIME[val] = False
facs = [idx for idx, val in enumerate(IS_PRIME) if val and idx >= 2]
cache = [-1] * N
cache[2] = cache[3] = 1
cache[1] = 0
for i in facs:
    cache[i] = 1
class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        n = len(nums)
        for num in nums:
            cp = num
            if cache[num] == -1:
                c = 0
                for fac in facs:
                    print(fac, num)
                    if fac * fac > num:
                        if num > 1:
                            c += 1
                        break
                    if num % fac == 0:
                        c += 1
                        while num % fac == 0:
                            num //= fac
                cache[cp] = c
        l = [-1] * n
        stack = []
        for idx, val in enumerate(nums):
            curr = cache[val]
            while stack and cache[nums[stack[-1]]] < curr:
                stack.pop()
            last = -1 if not stack else stack[-1]
            l[idx] = last
            stack.append(idx)
        r = [-1] * n
        stack = []
        for i in range(n - 1, -1, -1):
            val = nums[i]
            curr = cache[val]
            while stack and cache[nums[stack[-1]]] <= curr:
                stack.pop()
            last = n if not stack else stack[-1]
            r[i] = last
            stack.append(i)
        indices = sorted(range(n), key=lambda x: (nums[x], -x), reverse=True)
        # visited = SortedList()
        # visited.add(-1)
        # visited.add(n)
        ans = 1
        for idx in indices:
            # i = visited.bisect_left(idx)
            # li = max(visited[i - 1], l[idx])
            # lc = idx - li
            # ri = min(visited[i], r[idx])
            rc = (idx - l[idx]) * (r[idx] - idx)
            need = min(rc, k)
            # print(nums[idx], rc, cache[nums[idx]])
            ans = ans * pow(nums[idx], need, mod)
            ans %= mod
            k -= need
            if k == 0:
                break
            # visited.add(idx)
        return ans
43600