<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 computeSPF(UP):
    spf = [i for i in range(UP+1)]
    root = int(pow(UP,0.5))
    if(root * root <UP):
        root +=1
    for i in range(2,root+1):
        if(spf[i] != i):
            continue
        for j in range(i*i,UP+1,i):
            if(spf[j]>i):
                spf[j] = i
    return spf

def getDistinctPrimes(v,spf):
    pSet = set()
    while(v>1):
        pSet.add(spf[v])
        v //= spf[v]
    return len(pSet)

X = 10**5
spf = computeSPF(X)
#print(spf)

pScore = [0 for _ in range(X+1)]
for i in range(2,X+1):
    pScore[i] = getDistinctPrimes(i,spf)

#print(pScore)

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        MOD = 10**9 + 7
        n = len(nums)
        lt = [-1 for _ in range(n)]
        rt = [n for _ in range(n)]
        stack = []

        for i in range(n):
            s = pScore[nums[i]]
            while(stack and pScore[nums[stack[-1]]]<s):
                stack.pop()
            if(stack):
                lt[i] = stack[-1]
            stack.append(i)
        
        stack = []
        for i in range(n-1,-1,-1):
            s = pScore[nums[i]]
            while(stack and pScore[nums[stack[-1]]]<=s):
                stack.pop()
            if(stack):
                rt[i] = stack[-1]
            stack.append(i)
        # print(lt)
        # print(rt)
        values = []
        for i in range(n):
            cnt = (i-lt[i])*(rt[i]-i)
            values.append((nums[i],cnt))
        values.sort(key = lambda x:(-x[0]))

        ans = 1
        remain = k
        idx = 0
        while(remain>0):
            minCnt = min(remain,values[idx][1])
            temp = pow(values[idx][0],minCnt,MOD)
            ans *= temp
            ans %= MOD
            
            remain -= minCnt
            if(minCnt == values[idx][1]):
                idx +=1
            else:
                pass

        return ans%MOD
39700