Codeforces Round 768 (Div. 1) |
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Finished |
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:
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.
44 04 14 24 3
0 3 1 2 0 2 1 3 0 1 2 3 -1
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.
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