<p>You are given an array <code>nums</code> consisting of <strong>positive</strong> integers.</p>

<p>We call a subarray of <code>nums</code> <strong>nice</strong> if the bitwise <strong>AND</strong> of every pair of elements that are in <strong>different</strong> positions in the subarray is equal to <code>0</code>.</p>

<p>Return <em>the length of the <strong>longest</strong> nice subarray</em>.</p>

<p>A <strong>subarray</strong> is a <strong>contiguous</strong> part of an array.</p>

<p><strong>Note</strong> that subarrays of length <code>1</code> are always considered nice.</p>

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

<pre>
<strong>Input:</strong> nums = [1,3,8,48,10]
<strong>Output:</strong> 3
<strong>Explanation:</strong> The longest nice subarray is [3,8,48]. This subarray satisfies the conditions:
- 3 AND 8 = 0.
- 3 AND 48 = 0.
- 8 AND 48 = 0.
It can be proven that no longer nice subarray can be obtained, so we return 3.</pre>

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

<pre>
<strong>Input:</strong> nums = [3,1,5,11,13]
<strong>Output:</strong> 1
<strong>Explanation:</strong> The length of the longest nice subarray is 1. Any subarray of length 1 can be chosen.
</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>
</ul>

class Solution:
    def longestNiceSubarray(self, nums: List[int]) -> int:
        left=right=0
        ml=1
        l=len(nums)
        while right<l-1:
            if self.AND(nums,left,right,nums[right+1])==-1:
                right+=1
            else:
                left=self.AND(nums,left,right,nums[right+1])+1
            ml=max(ml,right-left+1)
        return ml
        

    def AND(self,array,left,right,k):
        for i in range(left,right+1):
            if array[i]&k!=0:
                return i
        return -1
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