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<bits/stdc++.h>
using namespace std;

#define ll long long
#define mx 2000005

vector <int> v;
   
void Print()
{
    for(auto it : v)
        cout << it << " ";
    cout << "\n";
}


void solve(int tc)
{
   v.clear();
   int n;
   cin >> n;
   for(int i=n; i>=1; i--){
       v.push_back(i);
   }

   if(n%2 == 0)
       Print();
   else{
       int mid = (n/2);
       int mid2 = (n/2)+1;

       swap(v[mid], v[mid2]);
       Print();
   }
   
}

int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    //calc();

    int t=1;
    cin >> t;
    
    for(int i=1; i<=t; i++)
        solve(i);

    return 0;
}