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>
//#include <cmath>
#define int long long
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL);cout.tie(NULL);
#define pb push_back
#define mod 1000000007
//#define lld long double
#define mii map<int, int>
#define mci map<char, int>
#define msi map<string, int>
#define pii pair<int, int>
#define ff first
#define ss second
#define bs(yup,x) binary_search(yup.begin(),yup.end(),x) //it will return bollean value
#define lb(yup,x) lower_bound(yup.begin(),yup.end(),x)  //it will return the adress of the number which is equal to or just greater than x ,and if it is equal to yup.end() than their in no such number exist
#define ub(yup,x) upper_bound(yup.begin(),yup.end(),x)
#define all(x) (x).begin(), (x).end()
#define rep(i,x,y) for(int i=x; i<y; i++)
#define fill(a,b) memset(a, b, sizeof(a))
#define vi vector<int>
#define setbits(x) __builtin_popcountll(x)
using namespace std;
const long long N=1000001, INF=2000000000000000000;
const long double EPS= 0.000000000001;
int power(int a, int b, int p)
    {
        if(a==0)
        return 0;
        int res=1;
        a%=p;
        while(b>0)
        {
            if(b&1)
            res=(res*a)%p;
            b>>=1;
            a=(a*a)%p;
        }
        return res;
    }
vi prime;
bool isprime[N];
void pre()
{
    for(int i=2;i<N;i++)
    {
        if(isprime[i])
        {
            for(int j=i*i;j<N;j+=i)
            isprime[j]=false;
            prime.pb(i);
        }
    }
    return;
}


int32_t main()
{
    
    IOS;
    //fill(isprime, true);
    //pre();
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        for(int i=2;i<=n;i++) cout<<i<<" ";
        cout<<"1\n";
    }
}