Mina was doing some research and during his research, he wrote down an array of non-negative integers $$$a_1, a_2, \dots, a_n$$$, and he calculated the value
$$$$$$x = a_1 \oplus a_2 \oplus \dots \oplus a_n$$$$$$
and wrote down the number $$$x$$$, where $$$\oplus$$$ denotes the "Bitwise XOR" (look at notes for more clarification).
However, his evil and distracting friend Ayman was teasing him, so he took exactly one index $$$j$$$ and erased $$$a_j$$$, and Mina didn't remember it, but the rest of the array and the number $$$x$$$ are still written down. Help Mina recover the value of $$$a_j$$$, so that he can proceed with his research.
The first line of input starts with an integer $$$t$$$ ($$$1 \le t \le 10^4$$$), the number of test cases.
The first line of each test case contains two integers $$$n$$$ and $$$x$$$ ($$$2 \le n \le 10^5$$$, $$$0 \le x \lt 2^{30}$$$, the size of the array and the number Mina wrote down).
The second line of each test case contains $$$n$$$ integers $$$a_1, a_2, \dots, a_n$$$ ($$$0 \le a_i \lt 2^{30}$$$, and there is exactly one $$$j$$$ such that $$$a_j = -1$$$, the integer that Ayman erased).
The sum of $$$n$$$ over all test cases doesn't exceed $$$5 \cdot 10^5$$$.
For each test case, output one line containing one integer, the value of $$$a_j$$$, where $$$j$$$ is the index such that $$$a_j = -1$$$ in the input.
It is guaranteed that $$$a_j$$$ satisfies $$$0 \le a_j \lt 2^{30}$$$, and it can be proven that there is only one unique integer $$$a_j$$$ satisfying the conditions given in the problem.
4 4 3 4 3 -1 1 5 6 7 1 -1 6 3 4 7 -1 1 2 4 2 0 1 -1
5 5 0 1
In the first test case, we note that $$$4 \oplus 3 \oplus 5 \oplus 1 = 3$$$, which is the given $$$x$$$.
The XOR of two bits $$$x$$$ and $$$y$$$ is $$$1$$$ if and only if $$$x \ne y$$$. For example $$$1$$$ XOR $$$0$$$ is $$$1$$$ and $$$1$$$ XOR $$$1$$$ is $$$0$$$. The bitwise XOR of any two integers $$$x$$$ and $$$y$$$ can be found by converting each integer into binary and taking the XOR of each two corresponding bits. For example, $$$5 \oplus 3 = 6$$$ since $$$3$$$ in binary is 011 and $$$5$$$ in binary is 101, so the result is 110 which is $$$6$$$.
