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 func(a,b):
      bin_a=bin(a)[2:]
      bin_b=bin(b)[2:]
      #print(a,b)
      max_len=len(bin_b)
      res1=[]
      b+=1
      for x in range(max_len):
          base=2**(x+1)
          #print(base)
          num=(a//base)*(base//2)+max(a%base-(base//2),0)
  #         print(num)
  #         print("--------")
          res1.append(num)
      
      res2=[]
      for x in range(max_len):
          base=2**(x+1)
          #print(base)
          num=(b//base)*(base//2)+max(b%base-(base//2),0)
  #         print(num)
  #         print("--------")
          res2.append(num)
      #print(res1,res2)
      final=[x2-x1 for x1,x2 in zip(res1,res2)]
      #print(final)
      max_num=max(final)
      print(b-a-max_num)
  
  t = int(input())
  while t > 0:
      t -= 1
      m, n = map(int, input().split())
      func(m, n)
  