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 <iostream>
#include <vector>

using namespace std;

int main()
{
	int cases=0;
	cin>>cases;
	vector<int> num;
	for(int i=0;i<cases;i++)
	{
		int temp=0;
		cin>>temp;
		num.push_back(temp);
	}
	vector< vector<int> > result;
	for(int i=0;i<cases;i++)
	{
		vector<int> temp(num[i]);
		if(num[i]%2==0)
		{
			for(int k=0;k<num[i];k++)
			{
				temp[(num[i]-1-k)] = k+1;
			}
		}
		else
		{
			for(int k=0;k<(num[i]-1);k++)
			{
				temp[k] = k+2;
			}
			temp[(num[i]-1)] = 1;
		}
		result.push_back(temp);
		for(int j=0;j<num[i];j++)
		{
			cout<<temp[j]<<" ";
		}
		cout<<endl;
	}
	return 0;
}