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;
int main()
{
    int t;
    scanf("%d",&t);
    for(int _ = 0;_<t;_++){
        int n;
        scanf("%d",&n);
        if(n%2==0){
            for(int i=n;i>0;i--){
                printf("%d ",i);
            }
        }
        else{
            vector<int> a(n);
            for(int i=n;i>0;i--){
                a[n-i]=i;
            }
            swap(a[n/2],a[0]);
            for(int i=0;i<n;i++){
                printf("%d ",a[i]);
            }
        }
        cout<<"\n";
    }
}