# AND Sorting solution codeforces

## AND Sorting solution codeforces

You are given a permutation pp of integers from 00 to n1n−1 (each of them occurs exactly once). Initially, the permutation is not sorted (that is, pi>pi+1pi>pi+1 for at least one 1in11≤i≤n−1).

The permutation is called XX-sortable for some non-negative integer XX if it is possible to sort the permutation by performing the operation below some finite number of times:

• Choose two indices ii and jj (1i<jn)(1≤i<j≤n) such that pi&pj=Xpi&pj=X.
• Swap pipi and pjpj.

Here && denotes the bitwise AND operation.

Find the maximum value of XX such that pp is XX-sortable. It can be shown that there always exists some value of XX such that pp is XX-sortable.

Input

The input consists of multiple test cases. The first line contains a single integer tt (1t104)(1≤t≤104)  — the number of test cases. Description of test cases follows.

The first line of each test case contains a single integer nn (2n2105)(2≤n≤2⋅105)  — the length of the permutation.

The second line of each test case contains nn integers p1,p2,...,pnp1,p2,…,pn (0pin10≤pi≤n−1, all pipi are distinct)  — the elements of pp. It is guaranteed that pp is not sorted.

It is guaranteed that the sum of nn over all cases does not exceed 21052⋅105.

Output

For each test case output a single integer — the maximum value of XX such that pp is XX-sortable.

Example

input

Copy
4
4
0 1 3 2
2
1 0
7
0 1 2 3 5 6 4
5
0 3 2 1 4


output

Copy
2
0
4
1

Note

In the first test case, the only XX for which the permutation is XX-sortable are X=0X=0 and X=2X=2, maximum of which is 22.

Sorting using X=0X=0:

• Contestants ranked 1st will win a Apple HomePod mini
• Contestants ranked 2nd will win a Logitech G903 LIGHTSPEED Gaming Mouse
• Contestants ranked 3rd ~ 5th will win a LeetCode Backpack
• Contestants ranked 6th ~ 10th will win a LeetCode water bottle
• Contestants ranked 11th ~ 20th will win a LeetCode Big O Notebook
• Swap p1p1 and p4p4p=[2,1,3,0]p=[2,1,3,0].
• Swap p3p3 and p4p4p=[2,1,0,3]p=[2,1,0,3].
• Swap p1p1 and p3p3p=[0,1,2,3]p=[0,1,2,3].

Sorting using X=2X=2:

• Swap p3p3 and p4p4p=[0,1,2,3]p=[0,1,2,3].

In the second test case, we must swap p1p1 and p2p2 which is possible only with X=0X=0.