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;
void swap(int &a, int& b){
	int t=a;
	a=b;
	b=t;
}
void solve(){
	long long int n,ans;
	scanf("%lld",&n);
	int p[n];
	for(int i=0;i<n;i++) p[i]=i+1;
	swap(p[0],p[n-1]);
	ans=(n-1)*(n-1);
	vector<pair<int,int>> v;
	v.push_back({1,n});
	for(int i=1;i<n/2;i++){
		ans+=2*(n-i-1)*(n-i-1);
		v.push_back({i+1,n});
		v.push_back({n-i,1});
		swap(p[0],p[n-i-1]);
		swap(p[n-1],p[i]);
	}
	if(n%2){
		v.push_back({n/2+1,1});
		ans+=(n/2)*(n/2);
		swap(p[n/2],p[0]);
	}
	reverse(v.begin(),v.end());
	printf("%lld\n",ans);
	for(int a:p) printf("%d ",a);
	puts("");
	printf("%lld\n",n-1);
	for(int i=0;i<n-1;i++) printf("%d %d\n",v[i].first, v[i].second);
}

int main(){
	int t;
	cin>>t;
	while(t--) solve();
}