Codeforces Round 869 (Div. 2) |
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Finished |
You're given a positive integer $$$n$$$.
Find a permutation $$$a_1, a_2, \dots, a_n$$$ such that for any $$$1 \leq l < r \leq n$$$, the sum $$$a_l + a_{l+1} + \dots + a_r$$$ is not divisible by $$$r-l+1$$$.
A permutation of length $$$n$$$ is an array consisting of $$$n$$$ distinct integers from $$$1$$$ to $$$n$$$ in arbitrary order. For example, $$$[2,3,1,5,4]$$$ is a permutation, but $$$[1,2,2]$$$ is not a permutation ($$$2$$$ appears twice in the array), and $$$[1,3,4]$$$ is also not a permutation ($$$n=3$$$ but there is $$$4$$$ in the array).
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 contain a single integer $$$n$$$ ($$$1 \leq n \leq 100$$$) — the size of the desired permutation.
For each test case, if there is no such permutation print $$$-1$$$.
Otherwise, print $$$n$$$ distinct integers $$$p_1, p_{2}, \dots, p_n$$$ ($$$1 \leq p_i \leq n$$$) — a permutation satisfying the condition described in the statement.
If there are multiple solutions, print any.
3 1 2 3
1 1 2 -1
In the first example, there are no valid pairs of $$$l < r$$$, meaning that the condition is true for all such pairs.
In the second example, the only valid pair is $$$l=1$$$ and $$$r=2$$$, for which $$$a_1 + a_2 = 1+2=3$$$ is not divisible by $$$r-l+1=2$$$.
in the third example, for $$$l=1$$$ and $$$r=3$$$ the sum $$$a_1+a_2+a_3$$$ is always $$$6$$$, which is divisible by $$$3$$$.
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