D. ConstructOR

You are given three integers $a$, $b$, and $d$. Your task is to find any integer $x$ which satisfies all of the following conditions, or determine that no such integers exist:

• $0 \leq x < {2}^{60}$;
• $a|x$ is divisible by $d$;
• $b|x$ is divisible by $d$.

Here, $|$ denotes the bitwise OR operation.

Input

Each test contains multiple test cases. The first line of input contains one integer $t$ $(1 \leq t \leq {10}^{4})$ — the number of test cases.

Each test case consists of one line, containing three integers $a$, $b$, and $d$ $(1 \leq a,b,d < {2}^{30})$.

Output

For each test case print one integer. If there exists an integer $x$ which satisfies all of the conditions from the statement, print $x$. Otherwise, print $−1$.

If there are multiple solutions, you may print any of them.

Example

input

8
12 39 5
6 8 14
100 200 200
3 4 6
2 2 2
18 27 3
420 666 69
987654321 123456789 999999999

output

18
14
-1
-1
0
11
25599
184470016815529983

Note

In the first test case, $x=18$ is one of the possible solutions, since $39|18=55$ and $12|18=30$, both of which are multiples of $d=5$.

In the second test case, $x=14$ is one of the possible solutions, since $8|14=6|14=14$, which is a multiple of $d=14$.

In the third and fourth test cases, we can show that there are no solutions.

解题思路

　　首先先确定无解的情况，对于$d$根据$\text{lowbit}$求出其二进制下末尾$0$的个数，假设为$k$。如果$a$或$b$在二进制下末尾$0$的个数小于$k$，那么一定无解。这是因为$d$含有因子$2^k$，而$a$或$b$包括按位或后的结果都不可能含有因子$2^k$，因此无解。

　　下面给出一种答案$x$的构造方案，让$a|x = b|x = x$，同时$x$也是$d$的倍数，这样至少可以保证$a|x$和$b|x$能被$d$整除（下面再证明$x$一定满足$x < 2^{60}$）。

　　由于$a, b < 2^{30}$，因此在第$0$到第$29$位中，如果$a$或$b$的第$i$位为$1$，那么$x$的第$i$为也应该要为$1$，这样就可以保证$a|x = b|x = x$。

　　那么如何保证$x$是$d$的倍数呢？当$x$的第$i$位要变成$1$时，我们直接让$x$加上$d$的$2^{i- k}$倍，即加上$2^{i-k} \times d$。$2^{i-k} \times d$等价于把$d$左移$i-k$位，这意味着$d$的最低位的$1$从第$k$位左移到了第$i$位，而此时$x$的第$i$位为$0$，因此$x$加上$2^{i-k} \times d$后第$i$位就变成了$1$，同时这样也保证了$x$始终是$d$的倍数。

　　下面来证明按照上面的构造方法得到的$x$一定是满足$x < 2^{60}$。考虑最糟糕的情况：$$\begin{align*} x &= 2^0 \times d + 2^1 \times d + \ldots + 2^{29} \times d \\ &= \frac{1 - 2^{30}}{1 - 2} \times d \\ &= \left( {2^{30} - 1} \right) \times d \\ &< 2^{30} \times 2^{30} = 2^{60} \end{align*}$$

　　AC代码如下：

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 typedef long long LL;
 5 
 6 int lowbit(LL x) {
 7     return x & -x;
 8 }
 9 
10 void solve() {
11     LL a, b, d;
12     scanf("%lld %lld %lld", &a, &b, &d);
13     LL ret = 0;
14     if (lowbit(a) < lowbit(d) || lowbit(b) < lowbit(d)) {   // a或b末尾0的个数小于d末尾0的个数
15         ret = -1;
16     }
17     else {
18         int k = log2(lowbit(d));    // d末尾0的个数
19         a |= b;
20         for (int i = k; i < 30; i++) {
21             // 如果a或b的第i位为1，并且此时x的第i位为0
22             if ((a >> i & 1) && (ret >> i & 1) == 0) ret += d << i - k; // 答案加上d * 2^(i-k)
23         }
24     }
25     printf("%lld\n", ret);
26 }
27 
28 int main() {
29     int t;
30     scanf("%d", &t);
31     while (t--) {
32         solve();
33     }
34     
35     return 0;
36 }

参考资料

　　Codeforces Round #833 (Div. 2) A - D：https://zhuanlan.zhihu.com/p/582927038

　　Codeforces Round #833 (Div. 2) A-D.md：https://www.cnblogs.com/BlankYang/p/16885937.html