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>
#ifndef ONLINE_JUDGE
#define debug(x) cout << #x << ": " << (x) << endl
#else
#define debug(x)
#endif
using namespace std;
typedef long long ll;
typedef vector<int> vi;
const int maxn=1e6+7,inf=0x3f3f3f3f,mod=1e9+7;

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    int kase;
    cin>>kase;
    while(kase--)
    {
        int n;
        cin>>n;
        vector<pair<int,int>>ans;
        if(n==2)
        {
            cout<<"1\n";
            cout<<"2 1\n";
            cout<<"1\n";
            cout<<"2 1\n";
        }
        else if(n==3)
        {
            cout<<"5\n";
            cout<<"2 3 1\n";
            cout<<"2\n";
            cout<<"1 3\n";
            cout<<"3 2\n";
        }
        else
        {
            ll sum=0;
            vi p(n+1);
            p[1]=n-1;
            for(int i=2;i<=n/2;++i) p[i]=i+1;
            p[n/2]=1;
            p[n/2+1]=n;
            for(int i=n/2+2;i<=n-1;++i) p[i]=i-1;
            p[n]=2;
            vi q=p;
            for(int i=2;i<=n/2;++i) ans.push_back({i,n}),assert(p[n]==i),swap(p[i],p[n]),sum+=1ll*(n-i)*(n-i);
            for(int i=n-1;i>=n/2+1;--i) ans.push_back({i,1}),assert(p[1]==i),swap(p[i],p[1]),sum+=1ll*(i-1)*(i-1);
            ans.push_back({1,n});
            sum+=1ll*(n-1)*(n-1);
            cout<<sum<<'\n';
            for(int i=1;i<=n;++i) cout<<q[i]<<' ';
            cout<<'\n';
            cout<<ans.size()<<'\n';
            for(auto &v:ans) cout<<v.first<<" "<<v.second<<'\n';
        }
    }
    return 0;
}