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 <bits/stdc++.h>
#include <vector>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#define ll long long
#define chushi(a, b) memset(a, b, sizeof(a))
#define endl "\n"
const double eps = 1e-8;
const ll INF=0x3f3f3f3f;
const int mod = 20000311;
const int maxn = 1e6 + 5;
const int N=1005;
using namespace std;

int v[maxn];
vector<pair<ll, ll> > res;

int main(){
	
	int ncase;
	cin >> ncase;
	while(ncase--){
		ll n;
		cin >> n;
		ll sum = 0;
		
		ll x = n/2;
		v[1] = n; v[n] = 1;
		res.push_back({n, 1});
		sum += (n-1) * (n-1);
		for(ll i = 1; i < n/2; i++){
			ll a = 1 + i;
			ll b = n - i;
			res.push_back({a, n});
			res.push_back({b, 1});
			v[a] = a; swap(v[a], v[n]);
			v[b] = b; swap(v[b], v[1]);
			sum += (n-a) * (n-a);
			sum += (n-a) * (n-a);
		}
		
		if(n & 1){
			v[x+1] = x+1;
			swap(v[1], v[x+1]);
			res.push_back({x+1, 1});
			sum += x * x;
		}
		
		cout << sum << endl;
		for(int i = 1; i <= n; i++) cout << v[i] << " ";
		cout << endl;
		n = res.size();
		cout << n << endl;
		for(int i = n-1; i >= 0; i--) cout << res[i].first << " " << res[i].second << endl; 
		res.clear();
	}
	

	return 0;
}