description: You are given an array consisting of all integers from [l, r] inclusive. For example, if l = 2 and r = 5, the array would be [2, 3, 4, 5]. What's the minimum number of elements you can delete to make the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of the array non-zero?
  
  A bitwise AND is a binary operation that takes two equal-length binary representations and performs the AND operation on each pair of the corresponding bits.
  
  Input
  
  The first line contains one integer t (1 ≤ t ≤ 10^4) — the number of test cases. Then t cases follow.
  
  The first line of each test case contains two integers l and r (1 ≤ l ≤ r ≤ 2 ⋅ 10^5) — the description of the array.
  
  Output
  
  For each test case, output a single integer — the answer to the problem.
  
  Example
  
  Input
  
  
  5
  1 2
  2 8
  4 5
  1 5
  100000 200000
  
  
  Output
  
  
  1
  3
  0
  2
  31072
  
  Note
  
  In the first test case, the array is [1, 2]. Currently, the bitwise AND is 0, as 1\ \& \ 2 = 0. However, after deleting 1 (or 2), the array becomes [2] (or [1]), and the bitwise AND becomes 2 (or 1). This can be proven to be the optimal, so the answer is 1.
  
  In the second test case, the array is [2, 3, 4, 5, 6, 7, 8]. Currently, the bitwise AND is 0. However, after deleting 4, 5, and 8, the array becomes [2, 3, 6, 7], and the bitwise AND becomes 2. This can be proven to be the optimal, so the answer is 3. Note that there may be other ways to delete 3 elements.
solution: 
  def nonZero(arr):
      l = len(bin(arr[1]))-2
      bit_count = [0]*(l)
      for i in range(l):
          zero = 0
          r = (arr[1] - arr[0])//(2**(i+1))
          zero += r*(2**i)
          idx1 = arr[0]%(2**(i+1))
          val1 = idx1//(2**(i))
          idx2 = arr[1]%(2**(i+1))
          val2 = idx2//(2**(i))
          # if i == l - 1:
          #     print(2**i)
          #     print(idx1, val1, idx2, val2)
          if val1 == 0 and val2 == 1:
              zero += (2**i - idx1)
          elif val2 == 0 and val1 == 1:
              zero += (idx2 + 1)
          elif val2 == val1:
              if val2 == 1:
                  if arr[1] - arr[0] > 2**i and idx2 < idx1:
                      zero += 2**i
              else:
                  if arr[1] - arr[0] > 2**i:
                      zero += 2**i - idx1 + idx2 + 1
                  else:
                      zero += arr[1] - arr[0] + 1
          bit_count[i] = zero
      return min(bit_count)
  
  r = int(input())
  res = []
  for i in range(r):
  
      arr = list(map(int, input().split()))
      res.append(nonZero(arr))
      
  for i in res:
      print(i)