Codeforces 1454A (A. Special Permutation) (DFS)

A. Special Permutation

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

You are given one integer nn (n>1n>1).

Recall that a permutation of length nn is an array consisting of nn distinct integers from 11 to nn in arbitrary order. For example, [2,3,1,5,4][2,3,1,5,4] is a permutation of length 55, but [1,2,2][1,2,2] is not a permutation (22 appears twice in the array) and [1,3,4][1,3,4] is also not a permutation (n=3n=3 but there is 44 in the array).

Your task is to find a permutation pp of length nn that there is no index ii (1≤i≤n1≤i≤n) such that pi=ipi=i (so, for all ii from 11 to nn the condition pi≠ipi≠i should be satisfied).

You have to answer tt independent test cases.

If there are several answers, you can print any. It can be proven that the answer exists for each n>1n>1.

Input

The first line of the input contains one integer tt (1≤t≤1001≤t≤100) — the number of test cases. Then tt test cases follow.

The only line of the test case contains one integer nn (2≤n≤1002≤n≤100) — the length of the permutation you have to find.

Output

For each test case, print nn distinct integers p1,p2,…,pnp1,p2,…,pn — a permutation that there is no index ii (1≤i≤n1≤i≤n) such that pi=ipi=i (so, for all ii from 11 to nn the condition pi≠ipi≠i should be satisfied).

If there are several answers, you can print any. It can be proven that the answer exists for each n>1n>1.

Example

input

Copy

2
2
5

output

Copy

2 1
2 1 5 3 4

//2021-03-11 09:25:03
#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

const int N = 1001;
int T, n;
bool vis[N];
int a[N];

bool dfs(int tot){
    if(tot == n+1){
        //printf("&&&& %d   
", tot);
        bool fff = 0;
        for(int i = 1; i <= n; i++){
            printf("%d ", a[i]);
        }printf("
");
        return true;
    }
    bool flag = 0;
//    printf("### %d   
", tot);
    for(int i = 1; i <= n; i++){
        if(!vis[i]){
            if(tot != i){
                a[tot] = i;
                //printf("a[%d] = %d
", tot, i);
            }
            else continue;
            vis[i] = 1;
            if(dfs(tot+1)){
                flag = 1;
                break;
            }
            vis[i] = 0;
        } 
    }
    if(flag) return true;
    else return false;
}


int main(){
    scanf("%d", &T);
    while(T--){
        scanf("%d", &n);
        memset(vis, 0, sizeof(vis));
        memset(a, 0, sizeof(a));
        dfs(1);
    }
    
    return 0;
}