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 ll long long int
#define pub push_back
#define vll vector<ll>
#define vpair vector<pair<ll,ll>>
#define All(X) (X).begin(),(X).end()
#define Reverse(X) reverse(All(X))
#define Unique(X) (X).erase(unique(All(X)),(X).end()))
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
using namespace std;
int main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    ll t;
    cin>>t;
    while(t--)
    {
        ll n;
        cin>>n;
        cout<<n<<" ";
        for(ll i=1;i<=n-1;i++)
            cout<<i<<" ";
        cout<<endl;
    }
    return 0;
}