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
  
  def solve():
  	n, m = map(int, input().split())
  	if n > m:
  		print(0)
  		return
  	for i in range(31):
  		c = (1<<i)
  		if c > n and c > m:
  			r = 0
  			for j in range(i,-1,-1):
  				if n ^ r > m:
  					print(r)
  					break
  				if ((n ^ r) | ((1<<(j+1))-1)) <= m:
  					r ^= (1<<(j+1))
  					print(r)
  					return
  				if (m & (1<<j)) != 0:
  					if (n & (1<<j)) == 0:
  						r |= (1<<j)
  						if n ^ r > m:
  							print(r)
  							return
  			if (n ^ r) <= m:
  				r ^= 1
  				print(r)
  				return
  			print(r)
  			break
  
  '''
  	for x in range(1000000000):
  		if n ^ x > m:
  			print(bin(n),bin(m),bin(x))
  			print(x)
  			return
  '''
  
  for i in range(int(input())):
  	solve()
  