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
   
  p2D = lambda x: print(*x, sep="\n")
  def II(): return int(sys.stdin.buffer.readline())
  def MI(): return map(int, sys.stdin.buffer.readline().split())
  def LI(): return list(map(int, sys.stdin.buffer.readline().split()))
  def LLI(rows_number): return [LI() for _ in range(rows_number)]
  def BI(): return sys.stdin.buffer.readline().rstrip()
  def SI(): return sys.stdin.buffer.readline().rstrip().decode()
  def li(): return [int(i) for i in input().split()]
  def lli(rows): return [li() for _ in range(rows)]
  def si(): return input()
  def ii(): return int(input())
  def ins(): return input().split()
  
  
  def solve(n,m):
      s=0
      util,pre=[],[]
      for i in range(31):
          if n&(1<<i)==0:
              s+=1<<i
              util.append(i)
              pre.append(s)
      diff=max(0,m-n+1)
      ans=0
      while(diff>0):
          for i in range(len(util)):
              if pre[i]>=diff:
                  ans|=(1<<util[i])
                  diff-=(1<<util[i])
                  break
      return ans
                  
  
  
  t=II()
  for _ in range(t):
      n,m=MI()
      print(solve(n,m))