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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: 1
Explanation: There is only one possible result: 0.

Example 2:

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

 

Constraints:

  • 1 <= nums.length <= 5 * 104
  • 0 <= nums[i] <= 109

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