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.
import sys
import math
import random
from queue import PriorityQueue as PQ
from bisect import bisect_left as BSL
from bisect import bisect_right as BSR
from collections import OrderedDict as OD
from collections import Counter
from itertools import permutations
# mod = 998244353
mod = 1000000007
MOD = mod
# sys.setrecursionlimit(1000000)
try:
    sys.stdin = open("actext.txt", "r")
    OPENFILE = 1
except:
    pass
 
def get_ints():
    return map(int,input().split())
 
def palindrome(s):
    mid = len(s)//2
    for i in range(mid):
        if(s[i]!=s[len(s)-i-1]):
            return False
    return True
def check(i,n):
    if(0<=i<n):
        return True
    else:
        return False

# -----------------------------------------------------------------------------------------

t = int(input())
for tt in range(t):
    n,k = get_ints()
    n = '{:032b}'.format(n)
    k = '{:032b}'.format(k+1)
    s = ''
    for i in range(32):
        if(k[i]=='1' and n[i]=='1'):
            s+='0'
        elif(k[i]=='1'):
            s+='1'
        elif(n[i]=='1'):
            break
        else:
            s+='0'
    s = s+'0'*(32-len(s))
    print(int(s,2))
        


