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: t=int(input())
  for _ in range(t):
      l,r=input().split(" ")
      l=int(l)
      r=int(r)
      length=r-l+1
      l-=1
      i=2
      ans=0
      while(r*2//i!=0):
          temp1=(l//i)*i/2
          if(l*2//i==0):
              temp1=0
          if(int(temp1)!=temp1):
              temp1=int(temp1+0.5)
          temp2=(r//i)*i/2
          if(int(temp2)!=temp2):
              temp2=int(temp2+0.5)
          #print(i,temp1,temp2)
          if(r%i!=0 and r*2>=i):
              temp=r-((r//i)*i+i/2)+1
              if(temp>0):
                  temp2+=temp
          if(l%i!=0 and l*2>=i):
              temp=l-((l//i)*i+i/2)+1
              if(temp>0):
                  temp1+=temp
          tep=temp2-temp1
          #print(i,temp1,temp2)
          if(tep>ans):
              ans=tep
          i=i*2
      print(int(length-ans))
  
  
  