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
  import random
  T=int(input())
  for t in range(T):
      #arrayz=random.sample(range(1,10000), 2)
      arrayz=[int(i) for i in input().split()]
      l=min(arrayz)
      r=max(arrayz)
      d=dict()
      s=1
      dig=[]
      for i in range(30):
          dig.append(s)
          s*=2
      mul=2
      for i in range(len(dig)):
          start=dig[i]
          if start>r:
              break
          m1=(l-start)/mul
          m1=math.ceil(m1)
          first=m1*mul+start
          if first<=r:
              sets=((r-first+1)//mul)
              tot=sets*(mul)/2
              nw=first+mul*sets
              tot+=max(min(r-nw+1,mul/2),0)
              gap=first-l
              gap-=(mul/2)
              tot+=max(0,gap)    
              d[i]=tot
          else:
              prev=first-mul
              tot=min(r+1,prev+mul/2)-max(prev,l)
              d[i]=tot
          mul*=2
  
      m=0
      for i in d:
          m=max(m,d[i])
      ans1=int(r-l+1-m)
      print(ans1)
      p='''d=dict()
      for i in range(l,r+1):
          s=bin(i)[2:]
          s=s[::-1]
          for i in range(len(s)):
              if s[i] =='1':
                  if i in d:
                      d[i]+=1
                  else:
                      d[i]=1
      m=0
      for i in d:
          m=max(m,d[i])
      ans2=(r-l+1-m)
      if ans1!=ans2:
          print("error",ans1,ans2,l,r)
      else:
          pass'''