<p>You are given a <strong>0-indexed</strong> integer array <code>nums</code> of length <code>n</code> and an integer <code>k</code>. In an operation, you can choose an element and multiply it by <code>2</code>.</p>

<p>Return <em>the maximum possible value of </em><code>nums[0] | nums[1] | ... | nums[n - 1]</code> <em>that can be obtained after applying the operation on nums at most </em><code>k</code><em> times</em>.</p>

<p>Note that <code>a | b</code> denotes the <strong>bitwise or</strong> between two integers <code>a</code> and <code>b</code>.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>

<pre>
<strong>Input:</strong> nums = [12,9], k = 1
<strong>Output:</strong> 30
<strong>Explanation:</strong> If we apply the operation to index 1, our new array nums will be equal to [12,18]. Thus, we return the bitwise or of 12 and 18, which is 30.
</pre>

<p><strong class="example">Example 2:</strong></p>

<pre>
<strong>Input:</strong> nums = [8,1,2], k = 2
<strong>Output:</strong> 35
<strong>Explanation:</strong> If we apply the operation twice on index 0, we yield a new array of [32,1,2]. Thus, we return 32|1|2 = 35.
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>1 &lt;= nums.length &lt;= 10<sup>5</sup></code></li>
	<li><code>1 &lt;= nums[i] &lt;= 10<sup>9</sup></code></li>
	<li><code>1 &lt;= k &lt;= 15</code></li>
</ul>

class Solution:
    def maximumOr(self, nums: List[int], k: int) -> int:
        if k == 0:
            maxx = max(nums)
            r = maxx*2**k
            ind = nums.index(maxx)
            ans = 0
            for i in range(len(nums)):
                if i == ind: continue
                ans = ans|nums[i]
            ans = ans|r
            return ans
        else:
            if len(nums) == 1: return 2**k*nums[0]
            if len(nums) == 2: return max(2**k*nums[0]|nums[1], nums[0]|nums[1]*2**k)
            ans = 0
            l = [nums[0]]
            r = [nums[-1]]
            for i in range(1,len(nums)):
                l.append(l[i-1]|nums[i])
            for n in nums[-2::-1]:
                r.append(r[-1]|n)
            r = r[-1::-1]
            maxx = 0
            for i in range(1,len(nums)-1):
                if maxx < (2**k*nums[i]|l[i-1]|r[i+1]):
                    maxx = (2**k*nums[i]|l[i-1]|r[i+1])
            if maxx < 2**k*nums[0]|r[1]: maxx = 2**k*nums[0]|r[1]
            if maxx < 2**k*nums[-1]|l[-2]: maxx = 2**k*nums[-1]|l[-2]
            return maxx




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