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 int             long long int
#define ff              first
#define ss              second
#define pb              push_back

#define vi              vector <int>
#define pii             pair <int, int>

#define loop(i,s,e)     for(int i=s;i<e;i++)
#define rloop(i,e,s)    for(int i=e;i>=s;i--)
#define mset(a,f)       memset(a,f,sizeof(a))

#define endl            "\n"
#define sz(x)           ((int) x.size())
#define all(p)          p.begin(), p.end()

#define bug(...)        __f (#__VA_ARGS__, __VA_ARGS__)
#define print(a)        for (auto x : a) cout << x << " "; cout << endl
#define Print(a,x,y)    for (int i=x; i<y; i++) cout<< a[i]<< " "; cout << endl



template <typename Arg1>
void __f (const char* name, Arg1&& arg1) { cout << name << " : " << arg1 << endl; }
template <typename Arg1, typename... Args>
void __f (const char* names, Arg1&& arg1, Args&&... args)
{
	const char* comma = strchr (names + 1, ',');
	cout.write (names, comma - names) << " : " << arg1 << " | "; __f (comma + 1, args...);
}




void solve()
{
	int n;
	cin >> n;
	vi ans;
	if (n % 2 == 0)
	{
		rloop(i, n, 1) ans.pb(i);
	}
	else
	{
		int x = n / 2;
		x++;
		ans.pb(x);
		rloop(i, n, 1)
		{
			if (i != x) ans.pb(i);
		}
	}
	print(ans);
}

int32_t main()
{
	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 t = 1;
	cin >> t;
	while (t--) solve();

	return 0;
}