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 st first
#define nd second
#define mp make_pair
#define pb push_back
#define N 1000005
#define M 2002
#define fre freopen("gir.gir","r",stdin),freopen("cik.cik","w",stdout)

#define pii pair<long long int,long long int>

typedef long long ll;
 
long long int x,y,z,w,n,m,k,flag,flag2,time2;



void solve(){
   cin>>n;
   cout<<n<<" ";
   for (int i = 1; i < n; ++i)
   {
        cout<<i<<" ";
   }
   cout<<endl;
}

int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    //fre;
    int TT;
    cin>>TT;
    while(TT--)solve();
    return 0;
}