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 collections import Counter
  from queue import Queue
  import math
  
  
  def func2(m, n):
      if m < (1 << n):
          return 0
      if m == (1 << n):
          return 1
  
      if m & (1 << n) != 0:
          k = (m | ((1 << n) - 1)) - ((1 << n) - 1)
          if k == (1 << n):
              return (m - k + 1)
          else:
              M = (m - k + 1) + (k - (1 << n)) // 2
              return M
      else:
          for k in range(n + 1, m):
              if m & (1 << k) != 0:
                  kk = (m | ((1 << k) - 1)) - (1 << k)
                  if kk >= m:
                      raise Exception
                  return func2(kk, n)
  
  
  if __name__ == '__main__':
      T = int(input())
  
      for t in range(T):
          l, r = [int(a) for a in input().split()]
  
          ans = r
          for n in range(r):
              if (1 << n) > r:
                  break
              ll, rr = func2(l - 1, n), func2(r, n)
              #print(ll, rr)
              ans = min(ans, (r - l + 1) - (rr - ll))
          print(ans)
  