A. NIT orz!time limit per test1 secondmemory limit per test512 megabytesinputstandard inputoutputstandard outputNIT, 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.InputEach 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.OutputFor each test case, print one integer — the answer to the problem.ExampleInput
52 33 45 50 2 4 6 81 9105 77 15 30 29 273 3954874310293834 10284344 13635445Output
7
13
11
31
48234367
NoteIn 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. 
52 33 45 50 2 4 6 81 9105 77 15 30 29 273 3954874310293834 10284344 13635445
7 13 11 31 48234367 
['bitmasks', 'greedy', '*800']