Problem: Lunar rover finally reached planet X. After landing, he met an obstacle, that contains permutation p of length n. Scientists found out, that to overcome an obstacle, the robot should make p an identity permutation (make p_i = i for all i).

Unfortunately, scientists can't control the robot. Thus the only way to make p an identity permutation is applying the following operation to p multiple times: 

  * Select two indices i and j (i ≠ j), such that p_j = i and swap the values of p_i and p_j. It takes robot (j - i)^2 seconds to do this operation. 

Positions i and j are selected by the robot (scientists can't control it). He will apply this operation while p isn't an identity permutation. We can show that the robot will make no more than n operations regardless of the choice of i and j on each operation.

Scientists asked you to find out the maximum possible time it will take the robot to finish making p an identity permutation (i. e. worst-case scenario), so they can decide whether they should construct a new lunar rover or just rest and wait. They won't believe you without proof, so you should build an example of p and robot's operations that maximizes the answer.

For a better understanding of the statement, read the sample description.

Input

The first line of input contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases.

Each of next t lines contains the single integer n (2 ≤ n ≤ 10^5) – the length of p.

Note, that p is not given to you. You should find the maximum possible time over all permutations of length n.

It is guaranteed, that the total sum of n over all test cases doesn't exceed 10^5.

Output

For each test case in the first line, print how many seconds will the robot spend in the worst case.

In the next line, print the initial value of p that you used to construct an answer.

In the next line, print the number of operations m ≤ n that the robot makes in your example.

In the each of next m lines print two integers i and j — indices of positions that the robot will swap on this operation. Note that p_j = i must holds (at the time of operation).

Example

Input


3
2
3
3


Output


1
2 1
1
2 1
5
2 3 1
2
1 3
3 2
5
2 3 1
2
1 3
2 3

Note

For n = 2, p can be either [1, 2] or [2, 1]. In the first case p is already identity, otherwise robot will make it an identity permutation in 1 second regardless of choise i and j on the first operation.

For n = 3, p can be equals [2, 3, 1].

  * If robot will select i = 3, j = 2 on the first operation, p will become [2, 1, 3] in one second. Now robot can select only i = 1, j = 2 or i = 2, j = 1. In both cases, p will become identity in one more second (2 seconds in total).
  * If robot will select i = 1, j = 3 on the first operation, p will become [1, 3, 2] in four seconds. Regardless of choise of i and j on the second operation, p will become identity in five seconds. 



We can show, that for permutation of length 3 robot will always finish all operation in no more than 5 seconds.

Solution (in CPP):

Solution:
#include <iostream>
#include <iomanip>
#include <vector>
#include <cmath>
#include <algorithm>
#include <set>
#include <utility>
#include <queue>
#include <map>
#include <assert.h>
#include <stack>
#include <string>
#define int long long
using namespace std;
/*
author: Legendary Grandmaster mshiladityam
Wowwwww. Loved it. What a beutiful problem. 
And the proof comes naturally with the solution.
Gr8 problem.

It's like: the best you can do for each index, actually exists, by construction.
*/

int sq(int x)
{
	return x*x;
}


void solve()
{
	int n;
	cin>>n;
	vector<int> v(n+1);
	
	v[1]=(n/2)+1;
	for (int j=n/2+1; j<n; j++)
	{
		v[j]=j+1;
	}
	v[n]=n/2;
	for (int j=n/2; j>1; j--) v[j]=j-1;
	
	
	int ans=0;
	for (int j=2; j<=n; j++) 
	{
		ans+=max(sq(j-1), sq(n-j));
	}
	cout<<ans<<"\n";
	for (int j=1; j<=n; j++) cout<<v[j]<<" ";
	cout << "\n";
	
	cout<<n-1<<"\n";
	
	for (int j=n/2+1; j<n; j++)
	{
		cout<<j<<" "<<1<<"\n";
	}
	for (int j=n/2; j>1; j--)
	{
		cout<<j<<" "<<n<<"\n";
	}
	cout<<1<<" "<<n<<"\n";
	return;
}

signed main()
{
	ios::sync_with_stdio(0);
	cin.tie(NULL); cout.tie(NULL);
	int t;
	cin >> t;
	//t=1;
	while (t--)
	{
		solve();
	}
	return 0;
}