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>
#define int long long
#define fast ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define M 1000000007
using namespace std;
signed main()
{
	fast
	int t;
	cin>>t;
	while(t--)
	{
		int n;
		cin>>n;
		int A[n], x = 0, a = 1, i = 0, j = n - 1;
		while(i <= j)
		{
			if(x % 2 == 0)
				A[j--] = a++;
			else
				A[i++] = a++;	
			x++;
		}	
		for(int i = 0 ; i < n ; i ++)
			if(A[i] == i + 1)
				swap(A[i], A[i-1]);
		
		for(int i = 0 ; i < n ; i ++)
			cout << A[i] << " ";
		cout << endl;
	}
}