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 math import ceil, log2, floor,sqrt,gcd
  from collections import defaultdict,deque
  from bisect import bisect_left
  import sys
  import os.path
  from itertools import permutations
  sys.setrecursionlimit(200000)
  if (os.path.exists('input.txt')):
      sys.stdin = open("input.txt", "r+")
      sys.stdout = open("output.txt", "w")
  
  
  # input = sys.stdin.readline
  # print=sys.stdout.write
  def MI():
      return map(int, input().split())
  
  mod=10**9+7
  
  
  def getcount(n, k):
      res = (n >> (k + 1)) << k 
      
      if ((n >> k) & 1):
          res += n & ((1 << k) -1)
          res+=1
      return res
  # print(getcount(14,2))
  def solve(testcase):
      # input()
      # n = int(input())
      # a=list(MI())
      l1,r1=MI()
      # c1=0
      # l1=l
      # r1=r
      # while(l>1):
      #     l=l//2
      #     c1+=1
      # c2=0
      # while(r>1):
      #     r=r//2
      #     c2+=1
  
      # print(c1,c2)
      ans=10**9
      t=(r1-l1+1)
      # print(t)
      for i in range(33):
          a=getcount(r1,i)
          b=getcount(l1-1,i)
          # print(a,b,i)
          ans=min(ans,t-(a-b))
  
      print(ans)
      # l=l1
      # r=r1
      # if c1==c2:
      #     print(0)
      # else:
      #     ans=10**9
      #     # print(r+1,2**(c1+1))
      #     ans=min(ceil((r+1-2**(c1+1))),ans)
      #     # print(ans)
  
      #     ans=min(ans, ((2**c2) - l))
  
      #     # ans= min(  ans,  )
      #     k= r+1-2**c2 
      #     c2-=1
      #     k+=2**c2 -l
      #     ans=min(ans,k)
      #     if l%2 and r%2:
      #         an=(r-l)//2
      #     else:
      #         an=(r-l+1)//2
      #     print(min(ans,an))
  
      
  
      
      
      
  t=int(input())
  # t=1
  for _ in range(t):
      solve(_+1)
  
  