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;

/*** Optimization ***/
#pragma GCC optimize("Ofast,no-stack-protector")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")

/*** Typedef ***/
typedef long long ll;
typedef unsigned long long ull;

/*** Input Output ***/
#define sci1(a) scanf("%d",&a)
#define sci2(a,b) scanf("%d %d",&a,&b)
#define sci3(a,b,c) scanf("%d %d %d",&a,&b,&c)

#define sc1(a) scanf("%lld",&a)
#define sc2(a,b) scanf("%lld %lld",&a,&b)
#define sc3(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)

#define pf1(a) printf("%lld\n",a)
#define pf2(a,b) printf("%lld %lld\n",a,b)

/*** Loops ***/
#define for0(num) for(ll i = 0; i < num; i++)
#define for1(num) for(ll i = 1; i <= num; i++)
#define forr(num) for(ll i = num - 1; i >= 0; i--)
#define forin(arr, num) for(ll i = 0; i < num; i++) sc1(arr[i]);

/*** Define Values ***/
#define mx 1000000
#define mod 1000000007
#define PI acos(-1.0)
#define eps 1e-7
#define size1 2005

#define pb push_back
#define ff first
#define ss second
#define mp make_pair
#define mem(name, value) memset(name, value, sizeof(name))

/*** STLs ***/
typedef vector <ll> vll;
typedef set <ll> sll;
typedef multiset <ll> msll;
typedef queue <ll> qll;
typedef map <ll, ll> mll;
typedef pair <ll, ll> pll;
typedef vector <pair <ll, ll> > vpll;

/*** Sorts ***/
#define all(v) (v).begin(), (v).end()
#define rev(v) reverse(all(v))
#define srt(v) sort(all(v))
#define srtGreat(v) sort(all(v), greater<ll>())

/*** Some Prints ***/
#define en cout << '\n';
#define no cout << "NO" << '\n'
#define yes cout << "YES" << '\n'
#define case cout << "Case " << t++ << ": "


int main()
{
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    vector<int> p(n);
    for (int i = 0; i < n; i++) {
      p[i] = n - i;
    }
    swap(p[n / 2], p[n - 1]);
    for (int i = 0; i < n; i++) {
      cout << p[i] << " ";
    }
    cout << endl;
  }
  return 0;
}