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:
//IN THE NAME OF GOD!
#include<bits/stdc++.h>

#define fileIO freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#define IOS ios::sync_with_stdio(0);cin.tie(0); cout.tie(0);
#define inf ((long long) 1e18 )
#define mod (1e9 + 7)
#define int long long
#define endl '\n'
#define pb push_back
#define eb emplace_back
#define all(x) (x).begin(),(x).end()
#define debug(x) cout << '>' << #x << ':' << x << '\n';
#define debug2(x,y) cout << '>' << #x << " = " << x << " & "<< #y <<" = "<< y << '\n';
using namespace std;

int32_t main() {
	IOS
#ifndef ONLINE_JUDGE
	fileIO
#endif
	auto start = std::chrono::high_resolution_clock::now();
	int test;
	test = 1;
	cin >> test;
	while (test--) {
		int n;
		cin >> n;
		vector<int> a(n);
		int j = 0;
		for (int i = n; i >= 1; i--)
			a[j++] = i;
		if (n & 1)
			swap(a[n / 2] , a[(n + 1) / 2]);
		for (auto i : a)
			cout << i << " ";
		cout << endl;
	}

	auto stop = std::chrono::high_resolution_clock::now();
	auto duration = std::chrono::duration_cast<std::chrono::nanoseconds>(stop - start);

	cerr << "Time taken : " << ((long double)duration.count()) / ((long double) 1e9) << "s " << endl;

	return 0;
}
// Author: YASHWANTH H L