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
  import math
  from collections import deque,Counter
  from sys import stdin
  
  #sys.setrecursionlimit(10**7)
  
  int1=lambda x: int(x)-1
  stinput=lambda :stdin.readline()[:-1]
  ii=lambda :int(stinput())
  mi=lambda :map(int, stdin.readline().split())
  li=lambda :list(mi())
  mi1=lambda :map(int1, stdin.readline().split())
  li1=lambda :list(mi1())
  mis=lambda :map(str, stdin.readline().split())
  lis=lambda :list(mis())
  
  from collections import defaultdict
  #d=defaultdict(int)
  #d=defaultdict(lambda:1)
  
  import heapq
  #heapq.heapify(hq)
  #heapq.heappop(hq)
  #heapq.heappush(hq,1)
  
  mod=10**9+7
  Mod=998244353
  INF=10**18
  ans=0
  
  def f(n):
    cnt=[0]*20
    for i in range(20):
      tmp=0
      m=(n)//(pow(2,i+1))
      tmp+=m*pow(2,i)
      k=m*pow(2,i+1)
      tmp+=max(n-k-pow(2,i)+1,0)
      cnt[i]=tmp
    return cnt
  
  t=ii()
  for _ in range(t):
    l,r=mi()
    n=r-l+1
    ans=n
    a=f(r)
    b=f(l-1)
    for i in range(20):
      ans=min(ans,n-(a[i]-b[i]))
    print(ans)