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"
using namespace std;

#define pb push_back
#define mp make_pair
#define fst first
#define snd second

#define fr(i,n) 	for(int i=0;i<n;i++)
#define frr(i,n)	for(int i=1;i<=n;i++)

#define ms(x,i)	memset(x,i,sizeof(x))
#define dbg(x)	cout << #x << " = " << x << endl
#define all(x)	x.begin(),x.end()

#define gnl cout << endl
#define olar cout << "olar" << endl
#define fastio ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)

typedef long long int ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<pii> vii;
typedef pair<ll,ll> pll;

const int INF = 0x3f3f3f3f;
const ll llINF = 0x3f3f3f3f3f3f3f;
const int MOD = 1e9+7;

int p[200200];

int main(){

	fastio;
	int T;
	cin >> T;
	vii ans;
	while(T--){
		ll n;
		cin >> n;
		ans.clear();
		frr(i,n) p[i] = i;
		ll resp = 0;

		frr(i, n - 1){
			if(i == 1){
				swap(p[1], p[n]);
				resp += (n - 1)*(n - 1);
				ans.pb(mp(1,n));
				continue;
			}

			ll dist = n - (i/2) - 1;
			if(i%2 == 0){
				swap(p[1], p[dist+1]);
				ans.pb(mp(dist + 1,1));
			}
			else{
				swap(p[n], p[n - dist]);
				ans.pb(mp(n - dist, n));
			}
			resp += dist*dist;
		}

		cout << resp << endl;
		frr(i,n) cout << p[i] << ' ';
		gnl;
		cout << n - 1 << endl;
		reverse(all(ans));
		for(auto x : ans){
			cout << x.fst << ' ' << x.snd << endl;
		}

	}

}