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: from collections import defaultdict as dd
  import math
  import sys
  input=sys.stdin.readline
  def nn():
      return int(input())
   
  def li():
      return list(input())
   
  def mi():
      return map(int, input().split())
   
  def lm():
      return list(map(int, input().split()))
   
  twos = [2**i for i in range(100)]
   
      
  
  def solve():
   
      l, r = lm()
      pows = 20
      l_counts = [0 for i in range(pows)]
      r_counts = [0 for i in range(pows)]
      for i in range(1,pows):
          l_counts[i] = l//(2**i)*(2**(i-1)) + max(0,l%(2**i)-2**(i-1))
      for i in range(1,pows):
          r_counts[i] = (r+1)//(2**i)*(2**(i-1)) + max(0,(r+1)%(2**i)-2**(i-1)) 
      #print(r,l)
      #print(r_counts)
      #print(l_counts)
      f_counts = [r_counts[i]-l_counts[i] for i in range(0,pows)]
      #print([r_counts[i]-l_counts[i] for i in range(0,pows)])
      print(r-l+1 - max(f_counts))
      #print("")
      
      
  q=nn()
  for _ in range(q):
      solve()