<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 sympy.ntheory import primefactors
class PrimeTable:
    def __init__(self, n:int) -> None:
        self.n = n
        self.primes = []
        self.max_div = list(range(n+1))
        self.max_div[1] = 1
        self.phi = list(range(n+1))

        for i in range(2, n + 1):
            if self.max_div[i] == i:
                self.primes.append(i)
                for j in range(i, n+1, i):
                    self.max_div[j] = i
                    self.phi[j] = self.phi[j] // i * (i-1)

    def is_prime(self, x:int):
        if x < 2: return False
        if x <= self.n: return self.max_div[x] == x
        for p in self.primes:
            if p * p > x: break
            if x % p == 0: return False
        return True
    def prime_factorization(self, x:int):
        if x > self.n:
            for p in self.primes:
                if p * p > x: break
                if x <= self.n: break
                if x % p == 0:
                    cnt = 0
                    while x % p == 0: cnt += 1; x //= p
                    yield p, cnt
        while (1 < x and x <= self.n):
            p, cnt = self.max_div[x], 0
            while x % p == 0: cnt += 1; x //= p
            yield p, cnt
        if x >= self.n and x > 1:
            yield x, 1

pt = PrimeTable(10 ** 5 + 1)

class Solution:
    def maximumScore(self, nums: List[int], k: int) -> int:
        
        tmp = []
        for x in nums:
            tmp.append(len(list(pt.prime_factorization(x))))
        n = len(nums)
        st = []
        n = len(tmp)
        alo, ahi = [-1] * n, [n] * n
        for i, x in enumerate(tmp):
            while st and tmp[st[-1]] < x:
                y = st.pop()
                ahi[y] = i
            if st: alo[i] = st[-1]
            st.append(i)
        fuck = [(i - alo[i]) * (ahi[i] - i) for i in range(len(tmp))]
        ev = []
        for i in range(n):
            ev.append((nums[i], fuck[i]))
        ev.sort(reverse=True)
        p = 0
        curr = 1
        MOD = int(1e9 + 7)
        #print(ev)
        while k > 0:
            x, freq = ev[p]
            d = min(k, freq)
            #for i in range(d):
            curr *= pow(x, d, MOD)
            curr %= MOD
            k -= d
            p += 1
        return curr % MOD
40200