We have an array A of non-negative integers.
For every (contiguous) subarray B = [A[i], A[i+1], ..., A[j]] (with i <= j), we take the bitwise OR of all the elements in B, obtaining a result A[i] | A[i+1] | ... | A[j].
Return the number of possible results.  (Results that occur more than once are only counted once in the final answer.)
Example 1:
Input: [0]
Output: 1
Explanation: 
There is only one possible result: 0.
Example 2:
Input: [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: [1,2,4]
Output: 6
Explanation: 
The possible results are 1, 2, 3, 4, 6, and 7.
Note:
1 <= A.length <= 50000
0 <= A[i] <= 10^9
import math
class Solution:
    def subarrayBitwiseORs(self, A: List[int]) -> int:
        res = set([])
        tmp = []
        maximpact = 0
        for i in A:
            newtmp = []
            newtmp_set = set([])
            maximpact = max(maximpact, int(math.log(i, 2)) + 1 if i > 0 else 0)
            if i not in newtmp_set:
                newtmp_set.add(i)
                newtmp.append(i)
                res.add(i)
            if maximpact > 0:
                for j in tmp:
                    k = i|j
                    if k not in newtmp_set:
                        newtmp_set.add(k)
                        newtmp.append(k)
                    res.add(k)
                    if len(newtmp) >= maximpact:
                        break
            tmp = newtmp
        return len(res)
            
                
            

import math
class Solution:
    def subarrayBitwiseORs(self, A: List[int]) -> int:
        res = set([])
        tmp = []
        maximpact = 0
        for i in A:
            newtmp = []
            newtmp_set = set([])
            maximpact = max(maximpact, int(math.log(i, 2)) + 1 if i > 0 else 0)
            if i not in newtmp_set:
                newtmp_set.add(i)
                newtmp.append(i)
                res.add(i)
            if maximpact > 0:
                for j in tmp:
                    k = i|j
                    if k not in newtmp_set:
                        newtmp_set.add(k)
                        newtmp.append(k)
                    res.add(k)
                    if len(newtmp) >= maximpact:
                        break
            tmp = newtmp
        return len(res)
            
                
            
