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 ll int long long
#define mp make_pair
//#define for(i,n) for(int i=0;i<n;i++)
#define F first
#define S second
#define fast ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define N 1000000007
#define pq priority queue
#define pb push_back
#define pf push_front
#define mt make_tuple
#define prec(a) fixed<<setprecision(9)<<a
#define endl "\n"
#define fast_set s.max_load_factor(0.25);s.reserve(500);
#define cy cout<<"YES"<<endl;
#define cn cout<<"NO"<<endl;
#define all(v) v.begin(),v.end()
#define allr(v) v.rbegin(),v.rend()
const long double PI = (long double)(3.1415926535897932384626433832795);

int main() 
{
	fast;
	ll test,n,i,j,k;
	cin>>test;
	while(test--)
	{
		cin>>n;
		ll ans=0;
		ll arr[n];
		arr[0]=n/2+1;
		arr[n-1]=n/2;
		for(i=1;i<n/2;i++)
		{
			arr[i]=i;
		}
		for(i=n/2;i<n-1;i++)
		{
			arr[i]=i+2;
		}
		vector<pair<ll,ll>>v;
		for(i=n/2;i<n-1;i++)
		{
			v.pb(mp(i+1,1));
		}
		for(i=n/2-1;i>=0;i--)
		{
			v.pb(mp(i+1,n));
		}
		for(auto it:v)
		{
			ans+=((it.F-it.S)*(it.F-it.S));
		}
		cout<<ans<<endl;
		for(i=0;i<n;i++)
		{
			cout<<arr[i]<<" ";
		}
		cout<<endl;
		cout<<v.size()<<endl;
		for(auto it:v)
		{
			cout<<it.F<<" "<<it.S<<endl;
		}
	}
	return 0;
}