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You are given an array `nums`

consisting of **positive** integers.

We call a subarray of `nums`

**nice** if the bitwise **AND** of every pair of elements that are in **different** positions in the subarray is equal to `0`

.

Return *the length of the longest nice subarray*.

A **subarray** is a **contiguous** part of an array.

**Note** that subarrays of length `1`

are always considered nice.

**Example 1:**

**Input:** nums = [1,3,8,48,10]
**Output:** 3
**Explanation:** 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.

**Example 2:**

**Input:** nums = [3,1,5,11,13]
**Output:** 1
**Explanation:** The length of the longest nice subarray is 1. Any subarray of length 1 can be chosen.

**Constraints:**

`1 <= nums.length <= 10`

^{5}`1 <= nums[i] <= 10`

^{9}

```
public int longestNiceSubarray(int[] A) {
int AND = 0, i = 0, res = 0, n = A.length;
for (int j = 0; j < n; ++j) {
while ((AND & A[j]) > 0)
AND ^= A[i++];
AND |= A[j];
res = Math.max(res, j - i + 1);
}
return res;
}
```

```
int longestNiceSubarray(vector<int>& A) {
int AND = 0, i = 0, res = 0, n = A.size();
for (int j = 0; j < n; ++j) {
while ((AND & A[j]) > 0)
AND ^= A[i++];
AND |= A[j];
res = max(res, j - i + 1);
}
return res;
}
```

```
def longestNiceSubarray(self, A):
res = AND = i = 0
for j in range(len(A)):
while AND & A[j]:
AND ^= A[i]
i += 1
AND |= A[j]
res = max(res, j - i + 1)
return res
```

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