description: You are given two integers n and m. Find the \operatorname{MEX} of the sequence n ⊕ 0, n ⊕ 1, …, n ⊕ m. Here, ⊕ is the [bitwise XOR operator](https://en.wikipedia.org/wiki/Bitwise_operation#XOR).
  
  \operatorname{MEX} of the sequence of non-negative integers is the smallest non-negative integer that doesn't appear in this sequence. For example, \operatorname{MEX}(0, 1, 2, 4) = 3, and \operatorname{MEX}(1, 2021) = 0. 
  
  Input
  
  The first line contains a single integer t (1 ≤ t ≤ 30 000) — the number of test cases.
  
  The first and only line of each test case contains two integers n and m (0 ≤ n, m ≤ 10^9).
  
  Output
  
  For each test case, print a single integer — the answer to the problem.
  
  Example
  
  Input
  
  
  5
  3 5
  4 6
  3 2
  69 696
  123456 654321
  
  
  Output
  
  
  4
  3
  0
  640
  530866
  
  Note
  
  In the first test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, 3 ⊕ 3, 3 ⊕ 4, 3 ⊕ 5, or 3, 2, 1, 0, 7, 6. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 4.
  
  In the second test case, the sequence is 4 ⊕ 0, 4 ⊕ 1, 4 ⊕ 2, 4 ⊕ 3, 4 ⊕ 4, 4 ⊕ 5, 4 ⊕ 6, or 4, 5, 6, 7, 0, 1, 2. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 3.
  
  In the third test case, the sequence is 3 ⊕ 0, 3 ⊕ 1, 3 ⊕ 2, or 3, 2, 1. The smallest non-negative integer which isn't present in the sequence i. e. the \operatorname{MEX} of the sequence is 0.
solution: from collections import deque
  import heapq
  from math import sqrt
  import operator
  from operator import itemgetter
  import sys
  
  import copy
  
   
  inf_var = 0
  if inf_var == 1:
      inf = open('input.txt', 'r')
  else:
      inf = sys.stdin
   
  input = inf.readline
  # sys.stdout = open('output.txt', 'w')
   
  def read_one_int():
      return int(input().rstrip('\n'))
  
  
  def read_one_float():
      return float(input().rstrip('\n'))
   
   
  def read_list_of_ints():
      res = [int(val) for val in (input().rstrip('\n')).split(' ')]
      return res
   
  def read_str():
      return input().rstrip()
  
  
  def main():
      tests = read_one_int()
  
      for _ in range(tests):
          n, m = read_list_of_ints()
  
          if m < n:
              print(0)
              continue
  
          if m == 0:
              print(n ^ (m + 1))
              continue
  
          n_bin = [0] * 30
          m_bin = [0] * 30
  
          bit = 1
          for i in range(30):
              if n & bit:
                  n_bin[i] = 1
              elif bit <= n:
                  n_bin[i] = 0
  
              if m & bit:
                  m_bin[i] = 1
              elif bit <= m:
                  m_bin[i] = 0
  
              bit <<= 1
  
          last_m_bit = None
          for i in range(len(m_bin) - 1, -1, -1):
              if m_bin[i] == 1:
                  last_m_bit = i
                  break
  
          bits_to_change = []
          can_decrease = False
          for i in range(last_m_bit - 1, -1, -1):
              if not can_decrease:
                  if m_bin[i]:
                      continue
                  elif n_bin[i]:
                      bits_to_change.append(i)
                      can_decrease = True
              else:
                  if m_bin[i] ^ n_bin[i]:
                      bits_to_change.append(i)
  
          
          if not bits_to_change:
              last_to_change = None
              for i in range(len(m_bin)):
                  if not m_bin[i] and not n_bin[i]:
                      bits_to_change.append(i)
                      last_to_change = i
                      break
  
              for i in range(last_to_change - 1, -1, -1):
                  if m_bin[i] ^ n_bin[i]:
                      bits_to_change.append(i)
  
          for bit_n in bits_to_change:
              val = 1 << bit_n
              m ^= val
  
          res = m ^ n
  
          print(res)
  
  
  if __name__== '__main__':
      main()
  