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 <algorithm>
#include <iostream>
#include <memory.h>
#include <iterator>
#include <cassert>
#include <fstream>
#include <cstdlib>
#include <vector>
#include <cstdio>
#include <string>
#include <queue>
#include <deque>
#include <cmath>
#include <ctime>
#include <stack>
#include <set>
#include <map>

using namespace std;

int main()
{
    cin.tie(0); ios_base::sync_with_stdio(0);
    int t;
    cin >> t;
    while (t--) {
        int n;
        cin >> n;
        if (n % 2 == 0) {
            for (int i = n; i >= 1; i--) cout << i << " ";
        }
        else {
            for (int i = n; i > n / 2 + 1; i--) cout << i << " ";
            cout << n / 2 + n % 2 - 1 << " " << n / 2 + n % 2 << " ";
            for (int i = n / 2 + n % 2-2; i >= 1; i--) cout << i << " ";
            
        }
        cout << "\n";
    }
}