Codeforces Round #640 (Div. 4) G.Special Permutation

Codeforces Round #640 (Div. 4) G.Special Permutation

題目鏈接
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.

Example

input

6
10
2
4
6
7
13

output

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 

簡單構造題~

• 對一個偶數，我們可以這樣輸出
  $2\ \ 4\ \ 6\ \ 8 \cdots n\ \ n-3\ \ n-1\cdots7\ \ 5\ \ 3\ \ 1$
• 對一個奇數，我們可以這樣輸出
  $1\ \ 3\ \ 5\ \ 7 \cdots n\ \ n-3\ \ n-1\cdots8\ \ 6\ \ 4\ \ 2$
  AC代碼如下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

main()
{
    int t,n;
    cin>>t;
    while(t--){
        cin>>n;
        if(n<=3) puts("-1");
        else{
            int flag=0;
            for(int i=1;i<=n;i++) if(i%2==n%2) cout<<i<<" ";
            for(int i=n;i>=1;i--){
                if(i%2!=n%2){
                    if(flag==0) cout<<i-2<<" ",flag++;
                    else if(flag==1) cout<<i+2<<" ",flag++;
                    else cout<<i<<" ";
                }
            }
            puts("");
        }
    }
}