Teza Round 1 (Codeforces Round 1015, Div. 1 + Div. 2)
A
Max and Mod
Codeforces
800
原题
←B →
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You are given an integer $n$. Find any permutation $p$ of length $n$$^{\text{∗}}$ such that:

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  • For all $2 \le i \le n$, $\max(p_{i - 1}, p_i) \bmod i$ $^{\text{†}}$ $= i - 1$ is satisfied.
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If it is impossible to find such a permutation $p$, output $-1$.

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$^{\text{∗}}$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). $^{\text{†}}$$x \bmod y$ denotes the remainder from dividing $x$ by $y$.

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Input

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Each test contains multiple test cases. The first line contains the number of test cases $t$ ($1 \le t \le 99$). The description of the test cases follows.

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The first line of each test case contains a single integer $n$ ($2 \le n \le 100$).

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Output

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For each test case:

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  • If such a permutation $p$ doesn't exist, output a single integer $-1$.
  • Otherwise, output $n$ integers $p_1, p_2, \ldots, p_n$ — the permutation $p$ you've found. If there are multiple answers, output any of them.
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Note

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In the first test case, it is impossible to find such a permutation $p$, so you should output $-1$.

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In the fourth test case, $p = [1, 5, 2, 3, 4]$ satisfies the condition:

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  • For $i = 2$, $\max(p_1, p_2) = 5$ and $5 \bmod 2 = 1$.
  • For $i = 3$, $\max(p_2, p_3) = 5$ and $5 \bmod 3 = 2$.
  • For $i = 4$, $\max(p_3, p_4) = 3$ and $3 \bmod 4 = 3$.
  • For $i = 5$, $\max(p_4, p_5) = 4$ and $4 \bmod 5 = 4$.
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样例

Input #1
4
2
3
4
5
Output #1
-1
3 2 1
-1
1 5 2 3 4
InputCopy OutputCopy
题目信息
题号
2084A
时间限制
1 s
内存限制
256 MB
标签
构造
数学

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