使用C++完成此题，语言标准C++11，给出完整，正确的代码。2s/512M.

题目描述

Noir has n gems. The i-th gem has a positive integer a_i written on it. Noir wants to divide these gems into several non-empty groups. Each gem belongs to exactly one group.

Suppose the i-th group contains gems with label $k{i,1}, k{i,2}, \ldots, k{i,p}$, then Noir treats the brightness of the i-th group as $a{k{i,1}} \oplus a{k{i,2}} \oplus \cdots \oplus a{k_{i,p}}, where \oplus$ is the bitwise-XOR operation. Denote the brightness of the i-th group as B_i.

For a grouping method of m groups, Noir treats the value of this method as B_1 \& B_2 \& \ldots \& B_m, where \& is the bitwise-AND operation.

Noir wants to find the maximum possible value over all grouping methods.

输入格式

The input contains multiple testcases. The first line of the input contains an integer T (1 \le T \le 10^4), the number of testcases.

For each testcase, the first line contains an integer n (1 \le n \le 5 \times 10^5), the number of gems.

The second line contains n integers a_1, a_2, \cdots, a_n (1 \le a_i < 2^{60}), the integers written on gems.

It’s guaranteed that the sum of n over all testcases does not exceed 5 \times 10^5.

输出格式

For each testcase, print an integer representing the maximum possible value over all grouping methods.

输入输出样例 #1

输入 #1

4
4
1 2 3 1
6
4 7 5 2 6 3
4
14 15 9 18
2
251508091405 13011908091815

输出 #1

2
6
26
13121614001578

说明/提示

For the first testcase, a possible grouping method is [1,2,3,1] = [1,3], [2,1] with a value of B_1 \& B_2 = (1 \oplus 3) \& (2 \oplus 1) = 2 \& 3 = 2. Another possible grouping method is [1,2,3,1] = [1,2,3,1], with a lower value B_1 = 1 \oplus 2 \oplus 3 \oplus 1 = 1. It can be proved that it’s not possible to achieve a value greater than 2.

For the second testcase, the best grouping method is [4,7,5,2,6,3] = [7], [5,3], [6], [4,2] with a value of 7 \& (5 \oplus 3) \& 6 \& (4 \oplus 2) = 6.