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:
// Problem Statement:  https://codeforces.com/contest/1474/problem/E/

/**
 *  Author: Ravi Kanth Gojur
 *  Code Forces: ravireddy07
 *  Code Chef: ravireddy115
 *  Github: ravireddy07
**/

#include <bits/stdc++.h>
#define ll long long int
#define ravireddy07 return
#define ii(a) scanf("%d", &a)
#define ii2(a, b) scanf("%d%d", &a, &b)
#define ii3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define ill(a) scanf("%lld", &a)
#define ill2(a, b) scanf("%lld%lld", &a, &b)
#define ill3(a, b, c) scanf("%lld%lld%lld", &a, &b, &c)
#define for1(i, a, b) for (int i = a; i < b; ++i)
#define for2(i, a, b) for (int i = b; i >= a; --i)
#define vi vector<int>
#define vii vector<vector<int>>
#define vl vector<ll>
#define vll vector<vector<ll>>
#define pii pair<int,int>
#define unmp unordered_map
#define pqi priority_queue<int>
#define pb push_back
#define sorta(a) sort(a.begin(), a.end())
#define sortd(a) sort(a.begin(), a.end(), greater<>())
#define sortr(a) sort(a.rbegin(), a.rend())
#define yes printf("YES\n")
#define no printf("NO\n")
using namespace std;

template <typename T, typename T1>
T amax(T &a, T1 b)
{
	if (b > a)
		a = b;
	return a;
}

const int N = 100100;
int n, m;
int p[N];
int ans[N][2];

ll sqr(ll x) {
	return x * x;
}


void harry() {
	ii(n);
	m = 0;
	for (int i = 1; i <= n; i++)
		p[i] = i;
	ll w = 0;
	int l = 1;
	int r = n - 1;
	int t = 0;
	while (m < n - 1) {
		if (t == 0) {
			w += sqr(n - l);
			ans[m][0] = l;
			ans[m][1] = n;
			m++;
			swap(p[l], p[n]);
			l++;
		} else {
			w += sqr(r - 1);
			ans[m][0] = r;
			ans[m][1] = 1;
			m++;
			swap(p[r], p[1]);
			r--;
		}
		t ^= 1;
	}

	printf("%lld\n", w);
	for (int i = 1; i <= n; i++)
		printf("%d ", p[i]);
	printf("\n");
	printf("%d\n", m);
	for (int i = m - 1; i >= 0; i--)
		printf("%d %d\n", ans[i][0], ans[i][1]);

	ravireddy07;
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
#endif

	int t;
	ii(t);
	while (t--)
		harry();
	return 0;
}