C. Interesting Sequence

Petya and his friend, robot Petya++, like to solve exciting math problems.

One day Petya++ came up with the numbers $n$ and $x$ and wrote the following equality on the board: $$n\ \&\ (n+1)\ \&\ \dots\ \&\ m = x,$$ where $\&$ denotes the bitwise AND operation. Then he suggested his friend Petya find such a minimal $m$ ($m \ge n$) that the equality on the board holds.

Unfortunately, Petya couldn't solve this problem in his head and decided to ask for computer help. He quickly wrote a program and found the answer.

Can you solve this difficult problem?

Input

Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 2000$). The description of the test cases follows.

The only line of each test case contains two integers $n$, $x$ ($0\le n, x \le 10^{18}$).

Output

For every test case, output the smallest possible value of $m$ such that equality holds.

If the equality does not hold for any $m$, print $-1$ instead.

We can show that if the required $m$ exists, it does not exceed $5 \cdot 10^{18}$.

Example

input

5
10 8
10 10
10 42
20 16
1000000000000000000 0

output

12
10
-1
24
1152921504606846976

Note

In the first example, $10\ \&\ 11 = 10$, but $10\ \&\ 11\ \&\ 12 = 8$, so the answer is $12$.

In the second example, $10 = 10$, so the answer is $10$.

In the third example, we can see that the required $m$ does not exist, so we have to print $-1$.

解题思路

　　首先遍历$n$和$x$在二进制下的每一位，用$n_k$来表示数字$n$二进制的第$k$位的数值，当遍历到第$k$位时发现$n_k = 0$，$x_k = 1$，那么就必定无解。因为整个过程都是按位与的运算，如果$n_k = 0$，那么无论与什么数相与得到的结果都是$0$而不可能得到$1$。

　　接下来分类讨论：

1. $n_k = 0$，$x_k = 0$。那么$m$可以取任何数，最终$n_k\ \&\ (n+1)_k\ \& \ \dots\ \&\ m_k = x_k = 0$总是成立。
2. $n_k = 1$，$x_k = 0$。为了按位与后第$k$位为$0$，那么很明显应该让$n$加上某个数$s$后$(n+s)_{k} = 0$，这样才能有$n_k\ \&\ (n+1)_k\ \&\ \dots\ (n+s)_k\ \&\ \dots\ \&\ m_k = x_k = 0$。很显然为了让$n+s$的第$k$位出现$0$，$s$可以取到无穷大，因此我们来看看$s$最小可以取到多少。可以发现$n$的第$k$位要进位时才由变$1$变$0$，因此只用考虑由$n$的前$k$位所构成的数，即$n\ \%\ {2^{k+1}}$，第$k$位进位后得到的结果为$2^{k+1}$，因此$s$的最小值为$2^{k+1} - n\ \%\ {2^{k+1}}$。
3. $n_k = 1$，$x_k = 1$。为了按位与后第$k$位始终为$1$，那么在$n \sim m$的数中不能出现第$k$位为$0$的数。一样的，当$n$的第$k$位要进位时才由变$1$变$0$，考虑由$n$的前$k$位所构成的数，即$n\ \%\ {2^{k+1}}$，那么在进位前且第$k$位仍为$1$所能取到的最大值是$2^{k+1}-1$，因此$s$的取值不应超过$2^{k+1}-1 - n\ \%\ {2^{k+1}}$。

　　然后对第$2$种情况的所有$s$取个最大值（即找到最大的下界）得到$l$，对第$3$种情况的所有$s$取个最小值（即找到最小的上界）得到$r$。如果$l \leq r$说明有解，那么答案就是$m = n + l$；否则就是无解。

　　AC代码如下：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 
 6 void solve() {
 7     LL n, m;
 8     scanf("%lld %lld", &n, &m);
 9     LL l = 0, r = 9e18;
10     for (int i = 0; i < 60; i++) {    // 1e18的二进制位最多有60位
11         LL x = n % (1ll << i + 1);    // 得到n的前i位所构成的数
12         int a = n >> i & 1, b = m >> i & 1;
13         if (!a && b) {    // n的第k位为0，m的第k位为1，无解
14             printf("-1\n");
15             return;
16         }
17         if (a && !b) l = max(l, (1ll << i + 1) - x);    // 第一种情况
18         else if (a && b) r = min(r, (1ll << i + 1) - 1 - x);    // 第二种情况
19     }
20     printf("%lld\n", l <= r ? n + l : -1);
21 }
22 
23 int main() {
24     int t;
25     scanf("%d", &t);
26     while (t--) {
27         solve();
28     }
29     
30     return 0;
31 }

参考资料

　　Codeforces Round #843 (Div. 2) C (思维)：https://zhuanlan.zhihu.com/p/598221402