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>
#include <unistd.h>
using namespace std;
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(), (x).end()
#define mid (l+r)/2
typedef long long ll;
typedef long double ld;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef pair<int, int> ii;

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