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 sys
  
  def readnum():
      input = sys.stdin.readline
      a=list(map(int,input().split()))
      return a
  
  def printf(val):
      sys.stdout.write(str(val) + '\n')
  
  def power(a, b, mod):
      if b == 1:
          return a
      if b == 0:
          return 1
  
      ans = power(a, b//2, mod)
      ans *= ans
  
      if b % 2:
          ans *= a
      
      ans = ans % mod
      return ans
  
  
  def factorial(n, start=1, skip=None):
      res = start
      for i in range(1, n+1):
          if i == skip:
              continue
          res *= i
  #        res = res % mod
  
      return res
  
  def make_deg(l):
      deg = {}
      for i in l:
          if not i in deg:
              deg[i] = 0
          deg[i] += 1
      return deg
  
  def get_num_of_bits(n):
      res = []
      for i in range(1, 31):
          res.append( 2**(i-1)*(n//(2**i)) + max(0, (n % (2**i)) - 2**(i-1)))
      return res
  
  def minus(l, r):
      res = []
      for i in range(len(l)):
          res.append(r[i] - l[i])
      return res
  
  for _ in range(readnum()[0]):
      l, r = readnum()
  #    print (l, get_num_of_bits(l+1))
  #    print (r, get_num_of_bits(r+1))
      res = minus(get_num_of_bits(l), get_num_of_bits(r+1))
  #    print (res)
      print (r - l - max(res) + 1)
  #    print ("" )
  