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;

#define LI long int
#define LL long long
#define VEC(T) vector<T>
#define PAIR(T_,T__) pair<T_,T__>
#define MOD 1000000007
#define INF 1000000000000000000ll

template <typename T>
void SWAP(T &_a, T &_b){ T t = _a;_a = _b; _b = t;}

template <typename T>
ostream &operator<<(ostream &stream, const vector<T>& inp_)
{ for(auto &x: inp_)stream << x << " "; stream << "\n";
return stream;}

template <typename T,typename T_>
ostream &operator<<(ostream &stream, const pair<T,T_>& inp_)
{stream << "(" << inp_.first << "," << inp_.second <<") ";
return stream;}

template <typename T>
T GCD(T a, T b)
{T temp;while(b>0){temp = b;b = a%b;a = temp;}
return a;}


int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);cout.tie(NULL);

	int t;
	cin >> t;
	while(t--) {
		int n;
		cin >> n;
		
		for(int i=n/2+1; i<=n; i++)
			cout << i << " ";
		for(int i=1; i<=n/2; i++)
			cout << i << " ";
		cout << "\n";
	}

	return 0;
}