Problem: You are given one integer n (n > 1).

Recall that 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 of length 5, 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).

Your task is to find a permutation p of length n that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).

You have to answer t independent test cases.

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

Input

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

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

Output

For each test case, print n distinct integers p_1, p_2, …, p_n — a permutation that there is no index i (1 ≤ i ≤ n) such that p_i = i (so, for all i from 1 to n the condition p_i ≠ i should be satisfied).

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

Example

Input


2
2
5


Output


2 1
2 1 5 3 4

Solution (in CPP):

Solution:
#include <iostream>
#include<unordered_map>
#include<stdio.h>
#include<bits/stdc++.h>
using namespace std;
long long int p=1e9+7;
#define endl '\n';
int main() {
   
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  t=1;
   cin>>t;
 
  while(t--)
  {
     
     
     
     
     
     
     int n;
     cin>>n;
     vector<int> v;
     for(int i=n;i>=1;i--)
     {
         v.push_back(i);
     }
     
     if(n%2==1)
     {
         int a=v[n/2];
         v[n/2]=v[n-1];
         v[n-1]=a;
     }
     
     for(int i=0;i<n;i++)
     {
         cout<<v[i]<<" ";
     }
     
     
     
     
     cout<<endl;
     
      
  }
  
  
  
  return 0;
       
      }