<p>You are given an array <code>nums</code> of <strong>non-negative</strong> integers and an integer <code>k</code>.</p>

<p>An array is called <strong>special</strong> if the bitwise <code>OR</code> of all of its elements is <strong>at least</strong> <code>k</code>.</p>

<p>Return <em>the length of the <strong>shortest</strong> <strong>special</strong> <strong>non-empty</strong> <span data-keyword="subarray-nonempty">subarray</span> of</em> <code>nums</code>, <em>or return</em> <code>-1</code> <em>if no special subarray exists</em>.</p>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,2,3], k = 2</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>

<p><strong>Explanation:</strong></p>

<p>The subarray <code>[3]</code> has <code>OR</code> value of <code>3</code>. Hence, we return <code>1</code>.</p>
</div>

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

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [2,1,8], k = 10</span></p>

<p><strong>Output:</strong> <span class="example-io">3</span></p>

<p><strong>Explanation:</strong></p>

<p>The subarray <code>[2,1,8]</code> has <code>OR</code> value of <code>11</code>. Hence, we return <code>3</code>.</p>
</div>

<p><strong class="example">Example 3:</strong></p>

<div class="example-block">
<p><strong>Input:</strong> <span class="example-io">nums = [1,2], k = 0</span></p>

<p><strong>Output:</strong> <span class="example-io">1</span></p>

<p><strong>Explanation:</strong></p>

<p>The subarray <code>[1]</code> has <code>OR</code> value of <code>1</code>. Hence, we return <code>1</code>.</p>
</div>

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

<ul>
	<li><code>1 &lt;= nums.length &lt;= 50</code></li>
	<li><code>0 &lt;= nums[i] &lt;= 50</code></li>
	<li><code>0 &lt;= k &lt; 64</code></li>
</ul>

class Solution:
    def countBit(self, n: int) -> int:
        bit_count = 0
        while n:
            bit_count += 1
            n >>= 1
        return bit_count

    def minimumSubarrayLength(self, nums: List[int], k: int) -> int:
        max_nums = max(nums)
        if max_nums >= k: return 1
        if (1 << self.countBit(max_nums)) <= k: return -1

        minlen = len(nums) + 1
        p1 = 0
        curr_or = nums[0]
        for p2 in range(1, len(nums)):
            curr_or |= nums[p2]
            if curr_or < k:
                continue
            curr_or = nums[p2]
            p1 = p2
            while p1 >= 0 and curr_or < k:
                p1 -= 1
                curr_or |= nums[p1]
            minlen = min(minlen, p2 - p1 + 1)
            p1 += 1
        return -1 if minlen > len(nums) else minlen
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