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 collections
  # import random
  # import math
  # import itertools
  # import math
  # import math
  #import bisect
  # from collections import defaultdict
  # import itertools
  # from sys import stdin, stdout
  # import math
  # import math
  # import bisect
  # import bisect
  # import collections
  # import itertools
  # import math
  # import collections
  # import itertools
  # import math
  import bisect
  import collections
  import math
  import sys
  
  # import operator
  # from decimal import Decimal
  
  # sys.setrecursionlimit(10**6)
  
  
  
  
  
  p2D = lambda x: print(*x, sep="\n")
  
  
  def II(): return int(sys.stdin.buffer.readline())
  
  
  def MI(): return map(int, sys.stdin.buffer.readline().split())
  
  
  def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
  
  
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  
  
  def BI(): return sys.stdin.buffer.readline().rstrip()
  
  
  def SI(): return sys.stdin.buffer.readline().rstrip().decode()
  
  
  def li(): return [int(i) for i in input().split()]
  
  
  def lli(rows): return [li() for _ in range(rows)]
  
  
  def si(): return input()
  
  
  def ii(): return int(input())
  
  
  def ins(): return input().split()
  
  
  # mod = 10**9+7
  # fval = [1]
  # for i in range(1,200001):
  #     x = (fval[-1]*i)%mod
  #     fval.append(x)
  #
  # ifac = [1]*(200001)
  # ifac[200000] = pow(fval[200000], -1, mod)
  # for i in range(200000-1, 1, -1): ifac[i] = ifac[i+1]*(i+1)%mod
  # _________________ For taking Input from the Text Files __________________
  # import os.path
  #
  # if (os.path.exists('input.txt')):
  #     sys.stdin = open("input.txt", "r")
  #     sys.stdout = open("output.txt", "w")
  #     sys.stderr = open("error.txt", "w")
  
  def drank(d, processing, da, rank):
      tmp = 10 ** 9
      if len(d[da]) == 1:
          return 1
      for di in d[da]:
          if processing[di - 1] == 0:
              processing[di - 1] = 1
              tmp = min(tmp, drank(d, processing, di, rank))
              processing[di - 1] = 0
      rank[da - 1] = tmp + 1
      return tmp + 1
  
  
  
  def binary_search(n, a):
      l = len(a)
      low = 0
      high = l-1
      while high>=low:
          mid = (high+low)//2
          if a[mid] == n:
              return mid
          elif a[mid]>n:
              high = mid-1
          else:
              low = mid+1
      return -1
  
  is_prime = [True]*(10**6+1)
  def primeseive(n):
      is_prime[0] = is_prime[1] = False
      i = 2
      while i*i<=n:
          if is_prime[i] == True:
              for j in range(i*i,n+1,i):
                  is_prime[j] = False
          i+=1
  #
  # m ,n = li()
  #     p = [li() for i in range(m)]
  #     print(p)
  #     a = []
  #     if m<n:
  #         for i in range(n):
  #             z = [p[j][i] for j in range(m)]
  #             a.append(max(z))
  #         return min(a)
  #     else:
  #         x = 0
  
  def solve():
      l , r = li()
      if l == r:
          return 0
      p = 1
      ans = 0
      while True:
          x = 2**p
          max_ones = ((r+1)//x)*(x//2)+ max((r+1)%x-x//2, 0)
          min_ones = ((l)//x)*(x//2)+ max((l)%x-x//2, 0)
          #print(max_ones)
          #print(min_ones)
          ans = max(ans, max_ones - min_ones)
          if x>r:
              break
          p+=1
      return (r-l+1)-ans
  
  
  
  def main():
      #primeseive(10**6)
      #print(is_prime)
  
      for _ in range(ii()):
          #si()
          #print(solve(), flush=True)
          sys.stdout.write(str(solve()) + "\n")
      #print(solve())
  
      #     z += str(ans) + '\n'
      # print(len(ans), ' '.join(map(str, ans)), sep='\n')
      # stdout.write(z)
  
  
  # for interactive problems
  # print("? {} {}".format(l,m), flush=True)
  # or print this after each print statement
  # sys.stdout.flush()
  
  
  if __name__ == "__main__":
      main()