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     FasterIO        ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define     bug(a,b)        cout<<a<<"\t"<<b<<"\n";
#define     Case(a,b)       cout<<"Case "<<a<<": "<<b<<"\n";
#define     precision(a,b)  fixed<<setprecision(a)<<b
#define     lp(i,a,b)       for(int i=a;i<b;i++)
#define     Endl            "\n"
#define     tab             "\t"
#define     reset(a,b)      memset(a,b,sizeof(a));
#define     sf(a)           scanf(" %d", &a);
#define     sfl(a)          scanf(" %lld", &a);
#define     srt(a)          sort(a.begin(),a.end());
#define     ALL(a)          a.begin(),a.end()
#define     pb(a)           push_back(a)
#define     pi              2*acos(0.0)
#define     yes             cout<<"YES\n";
#define     no              cout<<"NO\n";
#define     print2d(a,b,c)  for(int i =0; i <a; i++){for (int j = 0; j < b; j++){cout << c[i][j] << " ";}cout << "\n";}
#define     printAr(a,b)    for(int i =0; i <a; i++){cout<<b[i]<<" ";}cout << "\n";


typedef long long int ll;
typedef unsigned long long int ull;

const int sz = 1e5 + 9 , LOG = 20, infP = INT_MAX , infN = INT_MIN , mod = 1e9 + 7, eps = 1e-9;


void ini() {

    return;
}


void solve() {
    ll n , m = 1, k, l, r, ans = 0, sum = 0;
    cin >> n;
    int cur = n, ind = 1, mid = n / 2;
    for (int i = 1; i <= mid; i++)cout << cur-- << " ";
    cur = 1;
    for (int i = mid + 1; i <= n; i++)cout << cur++ << " ";
    cout << Endl;

}

int main() {
    FasterIO;
#ifndef ONLINE_JUDGE
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif
    // ini();
    int t = 1;
    cin >> t;
    while (t--)
        solve();
}