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:
//#pragma GCC optimize("ofast")
//#pragma GCC optimize("o3")
#include <bits/stdc++.h>
#define int long long
//#define ll long long
#define fr(i,x,y) for(int i=x;i<y;i++)
#define rfr(i,x,y) for(int i=x;i>=y;i--)
using namespace std;
const int N=1e7,M=1e2+1,INF=1e18,MOD=998244353,L=20,NN=1e9+1,PR=1e7;
bitset<PR> pr;
int rek(vector<int> p)
{
	int res=0;
	fr(i,0,p.size())
	{
		if(p[i]==i+1)continue;
		else 
		{
			int j=p[i]-1;
			int q=p[i]-i-1;
			q*=q;
			swap(p[i],p[p[i]-1]);
			res=max(res,rek(p)+q);
			swap(p[i],p[j]);
		}
	}
	return res;
}
void fnd(int n)
{
	vector<int> p(n);
	fr(i,0,n)
	{
		p[i]=i+1;
	}
	vector<int> ans;
	int c=0;
	while(next_permutation(p.begin(),p.end()))
	{
		int res=rek(p);
		if(res>c)
		{
			c=res;
			ans=p;
		}
	}
	cout<<c<<'\n';
	for(auto j:ans)cout<<j<<' ';
		cout<<'\n';
}
void solve()
{
	int n;
	cin>>n;
	int ans=0;
	vector<int> p(n);
	fr(i,0,n)p[i]=i+1;
	vector<pair<int,int> > op;
	swap(p[0],p[n-1]);
	op.push_back({n,1});
	ans+=(n-1)*(n-1);
	fr(i,1,n-1)
	{
		if(n-1-i>i)
		{
			swap(p[i],p[n-1]);
			op.push_back({i+1,n});
			ans+=(n-1-i)*(n-1-i);
		}
		else
		{
			swap(p[i],p[0]);
			ans+=i*i;
			op.push_back({i+1,1});
		}
	}
	cout<<ans<<'\n';
	fr(i,0,n)cout<<p[i]<<' ';
	cout<<'\n';
	reverse(op.begin(),op.end());
	cout<<op.size()<<'\n';
	for(auto j:op)cout<<j.first<<' '<<j.second<<'\n';
}
signed main()
{
	/*fr(i,2,PR)
	{
		if(pr[i])continue;
		for(int j=i+i;j<PR;j+=i)
		{
			pr[j]=1;
		}
	}*/
   //freopen("input.txt","r",stdin);
   //freopen("output.txt","w",stdout);
   //ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	/*int q;
	cin>>q;
	fnd(q);*/
   int t=1;
   cin>>t;
   while(t--)solve();
}
/*
25-3*3+2*2
3*3+2*2=13
1*1+2*2+2*2=8
 	
9  5
 ф
 
 
3*3*3
5*5
27+25
52
 
 
 
4*4+5*5+3*3+2*2

 
3 4 
1 5 5
*/