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;

typedef long long ll;
typedef long double ld;

typedef pair<int, int> pi;
typedef pair<ll,ll> pl;
typedef pair<ld,ld> pd;

typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<ll> vl;
typedef vector<pi> vpi;
typedef vector<pl> vpl;
typedef vector<vpl> vvpl;

#define REP2(i, a, b) for (int i=a; i<(b); i++)
#define REP(i, a) for (int i=0; i<(a); i++)
#define REPD(i, a) for (int i=a; i>=0; i--)
#define all(x) x.begin(),x.end()
#define S second
#define F first
#define PB push_back
#define POB pop_back

#define PRECISION(x) cout << fixed << setprecision(x)
#define FAST_IO ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

bool cp(int a,int b){
return a>b;
}

void solve(){
    int n;
    cin>>n;
    vi o,e;
    REP2(i,1,n+1){
    if(i%2){
        o.PB(i);
    }else{
        e.PB(i);
    }
    }
    int i=0,j=0,k=0;
    sort(all(o),cp);
    while(i<n){
        if(j<e.size()){
            cout<<e[j]<<" ";
            j++;
            i++;
        }
        if(k<o.size()){
            cout<<o[k]<<" ";
            k++;
            i++;
        }
    }
    cout<<endl;
}

int main(){
    FAST_IO;
    int t=1;
    cin>>t;
    while(t--){
        solve();

    }
    return 0;
}