<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 Prime(n):
    prime = [True] * n
    prime[0] = prime[1] = False
    ans = []
    
    for i in range(2, n):
        if prime[i]:
            ans.append(i)
        j = 0
        while j < len(ans) and i * ans[j] < n:
            prime[i * ans[j]] = False
            if i % ans[j] == 0:
                break
            j += 1
    return ans

primes = Prime(2 * 10 ** 5)
MOD = 10 ** 9 + 7
# print(primes)

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        
        keys = []
        for i, n in enumerate(nums):
            keys.append((n, i))  # sort by value, record indices
        
        keys.sort()
        
        
        # get the number of prime factors
        def isfj(a):
            a = abs(a)
            res = set()
            A = a
            x = 0
            
            while A and x < len(primes) and primes[x] * primes[x] <= a:
                while A % primes[x] == 0:
                    res.add(primes[x])
                    A = A // primes[x]
                x += 1
            
            #print("res", res)
            #print("remaining", A)
            
            if A == 1:
                return len(res)
            
            return len(res) + 1
            
    
        factors = []
        for n in nums:
            # print("factor num of ", n, " is ", isfj(n))
            factors.append(isfj(n))
        
        # print(factors)
        
        # find the coverage of each element to the left and right with monotonic st
        L, R = [], []
        
        st = []
        for i in range(len(factors)):
            n = factors[i]
            while st and factors[st[-1]] < n:
                st.pop()
            
            if st:
                L.append(i - st[-1])
            else:
                L.append(i + 1)

            st.append(i)
        
        
        factors = factors[::-1]
        
        st = []
        for i in range(len(factors)):
            n = factors[i]
            while st and factors[st[-1]] <= n:
                st.pop()
            
            if st:
                R.append(i - st[-1])
            else:
                R.append(i + 1)

            st.append(i)
        
        R = R[::-1]
        # factors = factors[::]
        
        #print("L", L)
        #print("R", R)
        
        # choose the best numbers
        res = 1
        while k > 0:
            n, i = keys.pop()
            comb = min(L[i] * R[i], k) # total possible combinations
            multiplier = pow(n, comb, MOD) % MOD
            res = res * multiplier
            k -= comb
        
        return res % MOD
            
35700