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: import math
  t = int(input())
  for test in range(t):
  	v = input().split(' ')
  	l = int(v[0])
  	r = int(v[1])
  	x1 = bin(l-1)
  	x2 = bin(r)
  	x1 = x1[2:]
  	x2 = x2[2:]
  	best = r-l
  	for i in range(max(len(x1),len(x2))):
  		v = (l-1)//(2**(i+1))
  		rem = (l-1)%(2**(i+1))
  		z = v*(2**i)
  		if(rem<2**i):
  			z+= rem+1
  		else:
  			z+= 2**i
  		v1 = r//(2**(i+1))
  		rem = r%(2**(i+1))
  		z1 = v1*(2**i)
  		if(rem<2**i):
  			z1 += rem+1
  		else:
  			z1 += 2**i
  		if(z1-z>=0):
  			best = min(best,z1-z)
  		
  
  		
  	print(best)