NIT orz!
Contest: Codeforces Global Round 21
Difficulty rating: 800
Tags: bitmasks, greedy

NIT, the cleaver, is new in town! Thousands of people line up to orz him. To keep his orzers entertained, NIT decided to let them solve the following problem related to $$$\operatorname{or} z$$$. Can you solve this problem too?

You are given a 1-indexed array of $$$n$$$ integers, $$$a$$$, and an integer $$$z$$$. You can do the following operation any number (possibly zero) of times:

- Select a positive integer $$$i$$$ such that $$$1\le i\le n$$$. Then, simutaneously set $$$a_i$$$ to $$$(a_i\operatorname{or} z)$$$ and set $$$z$$$ to $$$(a_i\operatorname{and} z)$$$. In other words, let $$$x$$$ and $$$y$$$ respectively be the current values of $$$a_i$$$ and $$$z$$$. Then set $$$a_i$$$ to $$$(x\operatorname{or}y)$$$ and set $$$z$$$ to $$$(x\operatorname{and}y)$$$.

Here $$$\operatorname{or}$$$ and $$$\operatorname{and}$$$ denote the bitwise operations OR and AND respectively.

Find the maximum possible value of the maximum value in $$$a$$$ after any number (possibly zero) of operations.

Input format:
Each test contains multiple test cases. The first line contains the number of test cases $$$t$$$ ($$$1 \le t \le 100$$$). Description of the test cases follows.

The first line of each test case contains two integers $$$n$$$ and $$$z$$$ ($$$1\le n\le 2000$$$, $$$0\le z<2^{30}$$$).

The second line of each test case contains $$$n$$$ integers $$$a_1$$$,$$$a_2$$$,$$$\ldots$$$,$$$a_n$$$ ($$$0\le a_i<2^{30}$$$).

It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^4$$$.

Output format:
For each test case, print one integer — the answer to the problem.

Constraints:
Time limit: 1.0 s
Memory limit: 512.0 MB

Note:
In the first test case of the sample, one optimal sequence of operations is:

- Do the operation with $$$i=1$$$. Now $$$a_1$$$ becomes $$$(3\operatorname{or}3)=3$$$ and $$$z$$$ becomes $$$(3\operatorname{and}3)=3$$$.
- Do the operation with $$$i=2$$$. Now $$$a_2$$$ becomes $$$(4\operatorname{or}3)=7$$$ and $$$z$$$ becomes $$$(4\operatorname{and}3)=0$$$.
- Do the operation with $$$i=1$$$. Now $$$a_1$$$ becomes $$$(3\operatorname{or}0)=3$$$ and $$$z$$$ becomes $$$(3\operatorname{and}0)=0$$$.

After these operations, the sequence $$$a$$$ becomes $$$[3,7]$$$, and the maximum value in it is $$$7$$$. We can prove that the maximum value in $$$a$$$ can never exceed $$$7$$$, so the answer is $$$7$$$.

In the fourth test case of the sample, one optimal sequence of operations is:

- Do the operation with $$$i=1$$$. Now $$$a_1$$$ becomes $$$(7\operatorname{or}7)=7$$$ and $$$z$$$ becomes $$$(7\operatorname{and}7)=7$$$.
- Do the operation with $$$i=3$$$. Now $$$a_3$$$ becomes $$$(30\operatorname{or}7)=31$$$ and $$$z$$$ becomes $$$(30\operatorname{and}7)=6$$$.
- Do the operation with $$$i=5$$$. Now $$$a_5$$$ becomes $$$(27\operatorname{or}6)=31$$$ and $$$z$$$ becomes $$$(27\operatorname{and}6)=2$$$.

Example input:
5
2 3
3 4
5 5
0 2 4 6 8
1 9
10
5 7
7 15 30 29 27
3 39548743
10293834 10284344 13635445

Example output:
7
13
11
31
48234367

Editorial:
Suppose we can only perform exactly one operation. In this case the answer is $$$S=\max_{1\le i\le n}(a_i\mathrm{\ or\ }z)$$$. In fact, we can prove that this is the answer.
Define $$$a_i'$$$ as the value of $$$a_i$$$ after some operations.
It suffices to prove the answer will never exceed $$$S$$$. Note that $$$z$$$ will always become a submask of itself after any number of operations, so $$$a_i$$$ will always be a submask of $$$(a_i\mathrm{\ or\ }z)$$$ after any number of operations. This leads to the conclusion that $$$a_i'\le (a_i\mathrm{\ or\ }z)$$$ for all $$$i$$$. Thus $$$\max_{1\le i\le n} a_i'\le \max_{1\le i\le n}(a_i\mathrm{\ or\ }z)=S$$$.
Time complexity is $$$O(n)$$$.