We have an array `arr`

of non-negative integers.

For every (contiguous) subarray `sub = [arr[i], arr[i + 1], ..., arr[j]]`

(with `i <= j`

), we take the bitwise OR of all the elements in `sub`

, obtaining a result `arr[i] | arr[i + 1] | ... | arr[j]`

.

Return the number of possible results. Results that occur more than once are only counted once in the final answer

**Example 1:**

Input:arr = [0]Output:1Explanation:There is only one possible result: 0.

**Example 2:**

Input:arr = [1,1,2]Output:3Explanation:The possible subarrays are [1], [1], [2], [1, 1], [1, 2], [1, 1, 2]. These yield the results 1, 1, 2, 1, 3, 3. There are 3 unique values, so the answer is 3.

**Example 3:**

Input:arr = [1,2,4]Output:6Explanation:The possible results are 1, 2, 3, 4, 6, and 7.

**Constraints:**

`1 <= nums.length <= 5 * 10`

^{4}`0 <= nums[i] <= 10`

^{9}

class Solution {
public int subarrayBitwiseORs(int[] arr) {
}
}