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.
from pickle import TRUE
import sys

input = sys.stdin.readline


############ ---- Input Functions ---- ############
def inp():
    return(int(input()))


def inlt():
    return(list(map(int, input().split())))


def insr():
    s = input()
    return(list(s[:len(s) - 1]))


def invr():
    return(map(int, input().split()))


def getHighestBitValue(v):
  return (1 << (v.bit_length() - 1))


def getPaddedBitString(v, bitStringLength):
  bitString = '{0:b}'.format(v)
  while len(bitString) < bitStringLength:
    bitString = '0' + bitString
  return bitString


def main():
  numTestCases = int(input())
  for _ in range(numTestCases):
    n, m = invr()
    # print("n=%s, m=%s" % (n, m))
    # nBitString = '{0:b}'.format(n)
    # mBitString = '{0:b}'.format(m)
    # bitStringLength = max(len(nBitString), len(mBitString))
    # print(getPaddedBitString(n, bitStringLength))
    # print(getPaddedBitString(m, bitStringLength))
    # print(getPaddedBitString(530866, bitStringLength))
    # print(getPaddedBitString(640, bitStringLength))
    
    if (n == 0):
      print(m + 1)
      continue
    
    if m == 0:
      print(0)
      continue
    
    highestNBit = getHighestBitValue(n)
    highestMBit = getHighestBitValue(m)
    
    if highestNBit > m:
      print(0)
      continue
    
    bitOn = highestMBit
    answerValue = m
    shutOffAll = False
    while bitOn > 0:
      bitOnInN = (bitOn & n) > 0
      bitOnInM = (bitOn & m) > 0
      
      if bitOnInN:
        if bitOnInM:
          # turn off this bit in answerValue
          answerValue &= ~bitOn
        else:
          # turn off all lower bits:
          answerValue &= ~(bitOn - 1)
          shutOffAll = True
          break
      bitOn >>= 1
    
    if not shutOffAll:
      answerValue += 1
    print(answerValue)


if __name__ == '__main__':
  main()
