Given a non-negative integer sequence $A=a_1,a_2,\ldots,a_n$ of length $n$, define

$$F(A)=\max_{1\le k<n}((a_1\mathbin{\&}a_2\mathbin{\&}\cdots\mathbin{\&}a_k)+(a_{k+1}\mathbin{\&}a_{k+2}\mathbin{\&}\cdots\mathbin{\&}a_n)).$$

where $\&$ is the bitwise-and operator.

You can perform the swapping operation at most once: choose two indices $i$ and $j$ such that $1\le i<j\le n$ and then swap the values of $a_i$ and $a_j$.

Calculate the maximum possible value of $F(A)$ after performing at most one swapping operation.

Input

There are multiple test cases. The first line of the input contains an integer $T$ indicating the number of test cases. For each test case:

The first line contains an integer $n$ ($2\le n\le 10^5$) indicating the length of sequence $A$.

The second line contains $n$ integers $a_1,a_2,\cdots,a_n$ ($0\le a_i\le 10^9$) indicating the given sequence $A$.

It’s guaranteed that the sum of $n$ of all test cases will not exceed $10^5$.

Output

For each test case output one line containing one integer indicating the maximum possible value of $F(A)$ after performing at most one swapping operation.

Examples

Input 1

3
6
6 5 4 3 5 6
6
1 2 1 1 2 2
5
1 1 2 2 2

Output 1

7
3
3

Note

For the first sample test case, we can swap $a_4$ and $a_6$ so the sequence becomes $\{6,5,4,6,5,3\}$. We can then choose $k=5$ so that $F(A)=(6\mathbin{\&}5\mathbin{\&}4\mathbin{\&}6\mathbin{\&}5)+(3)=7$.

For the second sample test case, we can swap $a_2$ and $a_4$ so the sequence becomes $\{1,1,1,2,2,2\}$. We can then choose $k=3$ so that $F(A)=(1\mathbin{\&}1\mathbin{\&}1)+(2\mathbin{\&}2\mathbin{\&}2)=3$.

For the third sample test case we do not perform the swapping operation. We can then choose $k=2$ so that $F(A)=(1\mathbin{\&}1)+(2\mathbin{\&}2\mathbin{\&}2)=3$.