题目描述

A permutation of length n is an array p=[p1,p2,…,pn], which contains every integer from 1 to n (inclusive) and, moreover, each number appears exactly once. For example, p=[3,1,4,2,5] is a permutation of length 5.

For a given number n (n≥2), find a permutation p in which absolute difference (that is, the absolute value of difference) of any two neighboring (adjacent) elements is between 2 and 4, inclusive. Formally, find such permutation p that 2≤|pi−pi+1|≤4 for each i (1≤i<n).

Print any such permutation for the given integer n or determine that it does not exist.

Input
The first line contains an integer t (1≤t≤100) — the number of test cases in the input. Then t test cases follow.

Each test case is described by a single line containing an integer n (2≤n≤1000).

Output
Print t lines. Print a permutation that meets the given requirements. If there are several such permutations, then print any of them. If no such permutation exists, print -1.

样例

6
10
2
4
6
7
13

9 6 10 8 4 7 3 1 5 2 
-1
3 1 4 2 
5 3 6 2 4 1 
5 1 3 6 2 4 7 
13 9 7 11 8 4 1 3 5 2 6 10 12 

如果你的n小于4 1~n中没有排列可以组成差值在二到四之间的 所以直接输出-1就好
在n大于4的时候 从后往前输出奇数因为最后输出的奇数一定是1 所以 只有 4 2 才可以满足 然后在让i从6开始输出偶数即可

#include<bits/stdc++.h>
using namespace std;
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);cout.tie(0);
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        if(n<4){
            cout<<-1<<endl;
            continue;
        }
        else{
            for(int i=n;i>=1;i--) if(i%2==1) cout<<i<<' ';
            cout<<4<<' '<<2<<' ';
            for(int i=6;i<=n;i++) if(i%2==0) cout<<i<<' ';
            cout<<endl;
        }
    }
    return 0;
}