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: import sys
  input = sys.stdin.buffer.readline
  
  # let ans be the smallest int not obtained from N ^ [0, M]
  # So, N ^ K = ans here K > M as we cover all xor from [0, M]
  # Now we find the min ans 
  # From max_bit to 0 if N(i) == K(i) then ans(i) = 0
  # if N(i), K(i) == (0, 1) then ans(i) = 1; for ans(i) = 0 we need K(i) = 0 but this dec K which we can't do
  # if N(i), K(i) == (1, 0) then ans(i) = 0; here we inc K(i) to 1 and since we increased this bit, all bits
  # from i-1 to 0 can be changed to match N(i) which will result in ans(i-1..0) = 0
  def solve():
      k = M + 1
      ans = 0
      for i in range(30, -1, -1):
          if (N >> i & 1) > (k >> i & 1): # N(i), K(i) == (1, 0)
              break
          ans |= (N & (1 << i)) ^ (k & (1 << i))
  
      return ans
  
  test_cases = int(input())
  for test_case in range(test_cases):
      N, M = map(int, input().split())
      print(solve())