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:
/*
    author:ujwalll
*/
#pragma GCC optimize("O2")
#include <bits/stdc++.h>
using namespace std; 
#define rep(i,n) for(int i=0;i<(n);i++)
#define for1(i,n) for(int i=1;i<=n;i++)
#define FOR(i,a,b) for(int i=(a);i<=(b);i++)
#define FORD(i,a,b) for(int i=(a);i>=(b);i--)
#define nx cout<<"\n";
const int INF = 1<<29;
const int MOD=1e9 + 7;
#define pp pair<ll,ll>
#define ppi pair<int,int>
typedef long long int ll;
#define pri(x) printf("%d ",x);
// #define pri(x) printf("%lld ",x);
bool isPowerOfTwo (ll x)  
{  
    return x && (!(x&(x-1)));  
}
void fastio()
{
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);
}
long long binpow(long long a, long long b) {
    long long res = 1;
    while (b > 0) {
        if (b & 1)
            res = res * a;
        a = a * a;
        b >>= 1;
    }
    return res;
}
const int dx[] = {1,0,-1,0,1,1,-1,-1};
const int dy[] = {0,-1,0,1,1,-1,-1,1};
////////////////////////////////////////////////////////////////////

int main(){
    fastio();
    int tc=1;
    cin>>tc;
    while(tc--){
        int n;
        cin>>n;
        int x=n;
        bool f=1;
        rep(i,n){
            if(n&1 && x==(n+1)/2 && f){
                cout<<1<<" ";
                f=0;
            }
            else
            cout<<x--<<" ";
        }
        cout<<"\n";
    }
    return 0;
}