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.readline
  
  T = int(input())
  for _ in range(T):
      n, m = map(int, input().split())
      answer = 0
      if n > m:
          print(0)
      else:
          nbin = bin(n)[2:]
          mbin = bin(m)[2:]
          nlen = len(nbin)
          mlen = len(mbin)
          nbin = '0' * (mlen-nlen) + nbin
          data = ['0'] * mlen
          flg = False
          #print(nbin)
          #print(mbin)
          for i in range(mlen):
              if nbin[i] == '0' and mbin[i] == '1':
                  data[i] = '1'
              elif nbin[i] == '1' and mbin[i] == '0':
                  data[i] = '0'
                  flg = True
                  break
          if not flg:
              #print(n ^ (m+1))
              flg2 = -1
              for i in range(mlen-1, -1, -1):
                  if nbin[i] == '0' and mbin[i] == '0':
                      data[i] = '1'
                      flg2 = i
                      break
              if flg2 >= 0:
                  for i in range(flg2+1, mlen):
                      data[i] = '0'
              else:
                  data = ['1'] + (['0'] * mlen)
              answer = int(''.join(data), 2)
              print(answer)
          else:
              #print(data)
              answer = int(''.join(data), 2)
              print(answer)
  