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:
/*** akshaykumar99 ***/
 
#include<bits/stdc++.h> 
using namespace std;
 
const long long MOD = 1e9+7;
const long long N = 1e5+1;
 
typedef long long ll;
typedef long double ld;

void test_case()
{
    ll n;
    cin >> n;
    for(int i = 1; i <= n; i++)
    {
    	cout << i % n + 1 << ' ';
    }
    cout << '\n';

}  
 
int32_t main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL); 
    int tc = 1;
    cin >> tc;
    while(tc--)
        test_case();
    return 0;
}