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 os
from io import BytesIO

# pypy2
sys.stdin = BytesIO(os.read(0, os.fstat(0).st_size))

f = sys.stdin

line = lambda: f.readline().strip('\r\n').split()

def write(w):
    sys.stdout.write(w)
    sys.stdout.write("\n")

def solve():
    if M <= N:
        return("0")
    
    res = 0
    m_bits = M.bit_length()
    n_bits = N.bit_length()
    for i in range(m_bits-n_bits):
        if M&(1<<(n_bits+i)):
            res |= 1<<(n_bits+i)
    for i in range(n_bits-1,-1,-1):
        m = M&(1<<i)
        n = N&(1<<i)
        if m and not n:
            res |= 1<<i
        elif not m and n:
            break
        
    return str(res)

T = int(line()[0])
for test in range(1,T+1):
    N,M = map(int,line())
    M += 1
    
    write(solve())
    
f.close()