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 pb push_back
#define ll long long

void solve()
{
    int n;
    cin>>n;
    int arr[n+1];

    for(int i=0;i<n;i++)
    {
        arr[i] = i+1;
    }

    int mid = n/2;
    
    if(n%2==0)
    {
    for(int i=n-1;i>=0;i--)
    {
        cout<<arr[i]<<" ";
    }
    cout<<endl;
    }
    else
    {
        for(int i=n-1;i>mid;i--)
        {
            cout<<arr[i]<<" ";
        }

        for(int i=0;i<=mid;i++)
        {
            cout<<arr[i]<<" ";
        }
        cout<<endl;
    }
}
int main(){
    int test;
    cin>>test;
    for(int i = 1; i <= test; i++){
        solve();
    }
return 0;
}