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:
#define _CRT_SECURE_NO_WARNINGS
#define _USE_MATH_DEFINES
#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <map>
#include <set>
#include <cmath>
#include <iomanip>
#include <queue>
#include <climits>
#include <stack>
#include <sstream>
#include <functional>
using namespace std;
typedef long long l64;

#define foor(i, n) for(int (i) = 0; (i) < (n) ; (i)++)
#define pb push_back

const bool multi_test = true;
int dd[22000];

bool check(vector<int> &a)
{
	int n = a.size();
	int ost = 0;
	foor(i, n)
	{
		if (ost > a[i])
		{
			return false;
		}
		else
		{
			ost = a[i] - ost;
		}
	}

	return ost == 0;
}

l64 sq(l64 s1, l64 s2)
{
	return (s1 - s2) * (s1 - s2);
}

bool solve()
{
	l64 n;
	cin >> n;

	vector<l64> a(n);
	foor(i, n)
		a[i] = i + 1;

	stringstream ss;
	vector<pair<l64, l64>> steps;

	l64 time = 0;
	l64 t;
	if(n % 2 == 1)
	{
		steps.emplace_back(1, n);
		time += sq(1, n);

		int mid = n / 2;
		steps.emplace_back(mid + 1, n);
		time += sq(mid + 1, n);

		for (int i = 1; i < mid; i++)
		{
			steps.emplace_back(mid + 1 - i, n);
			steps.emplace_back(mid + 1 + i, 1);
			time += sq(mid + 1 - i, n);
			time += sq(mid + 1 + i, 1);
		}
	}
	else
	{
		steps.emplace_back(1, n);
		time += sq(1, n);

		int mid = n / 2;
		for (int i = 0; i < mid - 1; i++)
		{
			steps.emplace_back(mid - i, n);
			steps.emplace_back(mid + 1 + i, 1);
			time += sq(mid - i, n);
			time += sq(mid + 1 + i, 1);
		}
	}

	foor(i, steps.size())
	{
		swap(a[steps[i].first - 1], a[steps[i].second - 1]);
	}

	cout << time << '\n';
	foor(i, n)
		ss << a[i] << ' ';

	ss << '\n' << steps.size();
	for(int i = steps.size() - 1; i >=0; i--)
		ss << '\n' << steps[i].first << ' ' << steps[i].second;

	cout << ss.str();



	return true;
}

int main()
{
#ifdef _DEBUG
	freopen("input.txt", "r", stdin);
#endif

	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	foor(i, 22000)
		dd[i] = 1;

	for (int i = 2; i < 22000; i++)
		if (dd[i] == 1)
			for (int j = i * i; j < 22000; j += i)
				dd[j] = 0;

	if (!multi_test)
	{
		solve();
		return 0;
	}

	int t;
	cin >> t;

	foor(i, t)
	{
		cin.ignore();
		if (!solve())
			return 0;

		cout << endl;
	}

	return 0;
}