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 $k_{i,1}, k_{i,2}, \dots, k_{i,p}$, 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 $B_1 \& B_2 \& \dots \& B_m$, where $\&$ is the bitwise-AND operation.

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

Input

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 $5 \times 10^5$.

Output

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

Example

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

2
6
26
13121614001578

Note

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 $7 \& (5 \oplus 3) \& 6 \& (4 \oplus 2) = 6$.