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 pii             pair<int,int>
#define vi              vector<int>
#define mii             map<int,int>
#define pqs             priority_queue<int,vi,greater<int> >
#define mod             1000000007
#define int             long long
#define pb              push_back
#define inf             1e18
#define sp(x,y)         fixed<<setprecision(y)<<x
#define mk(arr,n,type)  type *arr=new type[n];
#define w(t)            int t; cin>>t; while(t--)
#define mp              make_pair
#define endl            "\n"
#define pqb             priority_queue<int>
#define ceil(a , b)     ((a / b) + (a%b && 1))
#define rep(i,a,b)      for(int i=a;i<(int)b;i++)
#define repr(i,a,b)     for(int i=a;i>=(int)b;i--)
#define fastio          ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define setbits(x)      __builtin_popcountll(x)
#define zrobits(x)      __builtin_ctzll(x)
#define all(v)          v.begin(), v.end()
#define ff              first
#define ss              second
template<typename... T> void pn(T... args) { ((cout << args << " "), ...); cout<<"\n"; }
template<typename... T> void ps(T... args) { ((cout << args << " "), ...); cout<<""; }
// void start()
// {
//     ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
// #ifndef ONLINE_JUDGE
//     freopen("input.txt", "r", stdin);
//     freopen("output.txt", "w", stdout);
// #endif
// }
int power(int a, int b,int P=mod) {
    int res = 1;
    for (; b; b >>= 1, a = 1ll * a * a % P)
        if (b & 1)
            res = 1ll * res * a % P;
    return res;
}

void solve()
{
    int n;
    cin>>n;
    cout<<n<<" ";
    for(int i=1; i<n; i++)    cout<<i<<" ";
    cout<<endl;
}
int32_t main()
{
    fastio;
    w(t)
    {
        solve();
    }

    return 0;
}