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: T = int(input())
  for t in range(T):
      l, r = [int(i) for i in input().split()]
      diff = r-l+1
      i = 0
      ans = diff
      while (1<<i) <= r:
          
          at_r = (r // (2*(1<<i)) )*(1<<i)
          #print('i',i,'at_r',at_r)
          modval = 0
          if i >= 0:
              modval = r % (2*(1<<i))
          #print('modval',modval)
          incr = 0
          if modval >= (1<<i):
              incr = modval - (1<<i) + 1
          total_at_r = at_r + incr
          #print(2**i, 'total',total_at_r)
          #print('-'*50)
  
          at_l = ( (l-1) // (2*(1<<i)) )*(1<<i)
          modval = 0
          if i >= 0:
              modval = (l-1) % (2*(1<<i))
          #modval -= 1
          incr = 0
          if modval >= (1<<i):
              incr = modval - (1<<i) + 1
          total_at_l = at_l + incr
          #print(total_at_l, total_at_r,'for 2^i=',(1<<i))
          ans = min(ans, diff - (total_at_r - total_at_l))
          i += 1
          
      print(ans)
  
  '''
  0
  1
  2
  3
  4
  5
  6
  7
  8
  
  
  '''
  