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>

#define int long long int
#define mn 1e9+7
#define mx 0
#define pi 3.14159265358979323846
#define endl '\n'
#define loop(i, a, b) for (int i = a; i < b; i++)
#define vi vector<int>
#define vpi vector<pair<int, int>>
#define in(v) for (auto &i : v) cin >> i
#define all(v) v.begin(), v.end()
#define sz(a) a.size()
#define gif(a,b) ((a)%(b)==0?(a)/(b):(a)/(b)+1)
#define it(b) for(__typeof(b.begin())it=b.begin();it!=b.end();it++)

using namespace std;

int gcd(int a,int b){if(b==0)return a;return gcd(b,a%b);}

bool isPrime(int n){if(n<=1)return false;if(n<=3)return true;
	if(n%2==0||n%3==0)return false;for(int i=5;i*i<=n;i=i+6)
	if(n%i==0||n%(i+2)==0)return false;return true;}

int nCr(int n,int r){int res=1;if(n<r)return -1;r=min(r,n-r);
	loop(i,0,r){res*=(n-i);res/=(i+1);}return res;}

int nPr(int n,int r){int res=1;if(n<r)return -1;
	loop(i,0,r)res*=(n-i);return res;}


void shuruHoJaa()
{
	int n;
	cin>>n;
	loop(i, 0, n-1) cout<<i+2<<" ";
	cout<<1;


	return;
}

signed main()
{
	// #ifndef ONLINE_JUDGE
// 		freopen("input.txt","r",stdin);
// 		freopen("output.txt","w",stdout);
	// #endif
	std::ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);

	int t = 1;
	cin>>t;
	while(t--){ shuruHoJaa();if(t) cout<<endl;}

	cerr<<(1000*clock())/CLOCKS_PER_SEC<<"ms"<<endl;
	return 0;
}