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 _USE_MATH_DEFINES
using namespace std;
// #pragma GCC target ("avx2")
// #pragma GCC optimization ("O3")
// #pragma GCC optimization ("unroll-loops")
    
#define ll long long int
#define ld double
#define pb push_back
#define rep(i , j , n) for(ll i = j ; i < n ; i++)
#define pre(i , j , n) for(ll i = j ; i >= n ; i--) 
#define all(x) x.begin(), x.end()
typedef pair<int, int>	pii;
typedef pair<ll, ll>	pl;
typedef vector<int>		vi;
typedef vector<ll>		vll;
typedef vector<char>	vc;
typedef vector<bool>	vb;
typedef pair<ll,ll> 	pll;
#define br "\n"
#define ff first
#define ss second
#define gcd __gcd
#define debug(a) cout<<#a<<": "<<a;cout<<endl;
#define debugarr(a...) cout<<#a<<": ";for(auto it:a)cout<<it<<" ";cout<<endl;

void solve(){
    ll n;
    cin >> n;
    vll v(n);
    rep(i,1,n + 1){
        v[i - 1] = (i % n) + 1;
    }
    for(auto &x:v) cout << x << " ";
    cout << br;
}
 
 
int main(){
	ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    #ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #endif 
 
	ll t = 1;
    cin >> t;
	rep(i,0,t){
        // cout << "Case #" << i + 1 << ": ";
        solve();
        // test();
	}
 
}