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 os, sys
  
  isCoding = os.environ.get("CODE_ENV","")!=""
  if isCoding:
      sys.stdin = open('./examples/1.in', 'r')
  
  for _ in range(int(input())):
      l,r = map(int,input().split())
      n = len(bin(r)) - 2
      d = r-l+1
      a = [0]*20
  
      a[0] += d//2 + (d%2 * l%2)
  
      i = 1
      while i<20:
          t = 2**i
          zl = l//t
          if zl%2==0:
              zl+=1
          else:
              a[i] += min(r+1,(zl+1)*t) - l
              zl+=2
  
          zr = r//t
          if zr%2==0:
              zr-=1
  
          if zl<=zr:
              a[i] += t*((zr-zl)//2)
              zl = zr
              a[i] += min(r+1, (zl+1)*t) - zl*t
  
          i+=1
  
      a = [d - x for x in a]
      print(min(a))
  