CF1696A.NIT orz!

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.

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 $10^4$ .

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

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 $(27\operatorname{and}6)=2$ .