<p>We have an array <code>arr</code> of non-negative integers.</p> <p>For every (contiguous) subarray <code>sub = [arr[i], arr[i + 1], ..., arr[j]]</code> (with <code>i <= j</code>), we take the bitwise OR of all the elements in <code>sub</code>, obtaining a result <code>arr[i] | arr[i + 1] | ... | arr[j]</code>.</p> <p>Return the number of possible results. Results that occur more than once are only counted once in the final answer</p> <p> </p> <p><strong>Example 1:</strong></p> <pre> <strong>Input:</strong> arr = [0] <strong>Output:</strong> 1 <strong>Explanation:</strong> There is only one possible result: 0. </pre> <p><strong>Example 2:</strong></p> <pre> <strong>Input:</strong> arr = [1,1,2] <strong>Output:</strong> 3 <strong>Explanation:</strong> 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. </pre> <p><strong>Example 3:</strong></p> <pre> <strong>Input:</strong> arr = [1,2,4] <strong>Output:</strong> 6 <strong>Explanation:</strong> The possible results are 1, 2, 3, 4, 6, and 7. </pre> <p> </p> <p><strong>Constraints:</strong></p> <ul> <li><code>1 <= nums.length <= 5 * 10<sup>4</sup></code></li> <li><code>0 <= nums[i] <= 10<sup>9</sup></code></li> </ul>