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: # for _ in range(int(input())):
  #     n,k = map(int,input().split(" "))
  #     arr = list(map(int,input().split(" ")))
  #     ml = 0
  #     wa = []
  #     mx = -99999999999999999999
  #     if n>=100:
  #         for x in range(n-100,n):
  #             for y in range(x+1,n):
  #                 b = (x+1)*(y+1) - k*(arr[x]|arr[y])
  #                 if b >mx:
  #                     mx = b
  #     else:
  #         for x in range(n):
  #             for y in range(x+1,n):
  #                 b = b = (x+1)*(y+1) - k*(arr[x]|arr[y])
  #                 if b >mx:
  #                     mx = b
  #     print(mx)
  
  for _ in range(int(input())):
      n,m = map(int,input().split(" "))
  
      # Here we have to satisfy n ^ k >= (m+1)
      # and value of k must be as minimum as possible
  
      p = m +1
      p = bin(p)[2:]
      z = bin(n)[2:]
  
      if len(z)>=len(p):
          p = p.zfill(len(z))
          # print("hel1")
      else:
          z=z.zfill(len(p))
          # print(len(p))
          # print("hel2")
      k = ""
  
      # print("z =",  z)
      # print("p =",  p)
      for x in range(len(z)):
          if z[x]=='0' and  p[x]=='0':
              k+='0'
          elif z[x]=='0' and  p[x]=='1':
              k+='1'
          elif z[x]=='1' and  p[x]=='0':
              for y in range(x,len(z)):
                  k += '0'
              break
          else:
              k+='0'
      print(int(k,2))
          