A14798. And Matching

You are given a set of $n$ ( $n$ is always a power of $2$ ) elements containing all integers $0, 1, 2, \ldots, n-1$ exactly once.

Find $\frac{n}{2}$ pairs of elements such that:

- Each element in the set is in exactly one pair.
- The sum over all pairs of the [bitwise AND](https://en.wikipedia.org/wiki/Bitwise_operation#AND) of its elements must be exactly equal to $k$ . Formally, if $a_i$ and $b_i$ are the elements of the $i$ -th pair, then the following must hold: $$$$\sum_{i=1}^{n/2}{a_i \& b_i} = k, $$ where $\\&amp;$ denotes the bitwise AND operation. </li></ul><p>If there are many solutions, print any of them, if there is no solution, print $-1$$$ instead.

The input consists of multiple test cases. The first line contains a single integer $t$ ( $1 \leq t \leq 400$ ) — the number of test cases. Description of the test cases follows.

Each test case consists of a single line with two integers $n$ and $k$ ( $4 \leq n \leq 2^{16}$ , $n$ is a power of $2$ , $0 \leq k \leq n-1$ ).

The sum of $n$ over all test cases does not exceed $2^{16}$ . All test cases in each individual input will be pairwise different.

For each test case, if there is no solution, print a single line with the integer $-1$ .

Otherwise, print $\frac{n}{2}$ lines, the $i$ -th of them must contain $a_i$ and $b_i$ , the elements in the $i$ -th pair.

If there are many solutions, print any of them. Print the pairs and the elements in the pairs in any order.

In the first test, $(0\&3)+(1\&2) = 0$ .

In the second test, $(0\&2)+(1\&3) = 1$ .

In the third test, $(0\&1)+(2\&3) = 2$ .

In the fourth test, there is no solution.