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 pb push_back
#define f first
#define s second
#define go ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
/*

ll dx[8]={-1,-1,-1,0,0,1,1,1};
ll dy[8]={-1,0,1,-1,1,-1,0,1};
bool sortbysec(const pair<int,int> &a,
              const pair<int,int> &b)
{
    return (a.second < b.second);
}
*/
int main()
{
     ll t;
     cin>>t;
     while(t--)
     {
         ll n;
         cin>>n;
         for(int i=2;i<=n;i++)
            cout<<i<<" ";
         cout<<"1"<<endl;
     }

}