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 ll long long 
#define ld long double
#define llinf 1e17
#define nllinf -4e18
 
#define vll vector<ll>
#define sll set<ll>
#define pll pair<ll,ll> 
#define vpll vector<pair<ll,ll > >
 
#define repup(a,b,i) for(i=a;i<=b;i++)
#define repdown(a,b,i) for(i=b; i>=a; i--)
 
#define pb push_back
#define pf push_front
#define mp make_pair
#define fr first
#define sc second
 
#define err(n) cout<<n<<" ";
#define errp(a,b) cout<<"("<<a<<","<<b<<')'<<" ";
#define t1 cout<<",";

ll MOD = 1;

void add_self (ll& a, ll b){
    a+= b;
    a%=MOD;
}

void sub_self (ll& a, ll b){
    a-=b;
    a= ((a%MOD)+MOD)%MOD;
}

ll mul(ll a, ll b){
    a*=b;
    a%=MOD;
    return a;
}

void min_self(double& a, double b){
    a= min(a,b);
}

vpll g[100005];
bool visited[100005];
ll dist[100005];

int main(){
ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
ll i,j;
ll t; cin>>t;
while(t){
	t--;
	ll n;cin>>n;
	for(int i=1; i<=n; i++){
		cout<<(i)%n +1<<" ";
	}
	cout<<'\n';
}
}