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() {

#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	// solution comes here
	int numTests, n;
    cin >> numTests;
    vector<int> o;
    for (int i = 1; i <= numTests; i++) {
    	cin >> n;
    	for (int j = 0; j < n; j++) {
    		o.push_back(n - j);
    	} 
    	if (n % 2 == 1) {
    		int temp = o[n / 2];
    		o[n / 2] = o[0];
    		o[0] = temp;
    	}
    	for (auto it = o.begin(); it != o.end(); it++) {
        	cout << *it << " "; 
    	}
    	cout << '\n';
    	o.clear();
    }
	return 0;
}