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(x) push_back(x)
    #define FOR(i,a,b) for(int i=a;i<b;i++)
    typedef long long ll;

    int main(){
        ios::sync_with_stdio(false);
        cin.tie(0);

        int t;
        cin>>t;
        while(t--){
            int n;
            cin>>n;
            vector<int> v1;
            stack<pair<int,int> > s;
            FOR(i,0,n) v1.pb(i+1);
            int left=0;
            int right=n-1;
            ll tc=0;
            ll nc=0;
            swap(v1[left],v1[right]);
            tc+=pow(right-left,2);
            nc+=1;
            s.push(make_pair(left+1,right+1));
            while(left<right-1){
                right-=1;
                swap(v1[0],v1[right]);
                tc+=pow(right,2);
                nc++;
                s.push(make_pair(right+1,1));
                if(left<right-1){
                    left+=1;
                    swap(v1[left],v1[n-1]);
                    tc+=pow(n-1-left,2);
                    nc++;
                    s.push(make_pair(left+1,n));
                }
            }
            cout<<tc<<endl;
            FOR(i,0,n) cout<<v1[i]<<" ";
            cout<<endl;
            cout<<nc<<endl;
            while(!s.empty()){
                pair<int,int> p=s.top();
                s.pop();
                cout<<p.first<<" "<<p.second<<endl;
            }

        }

        return 0;
    }