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 int long long
#define endl '\n'

signed main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);

    int t;
    cin >> t;

    for(int nr = 0; nr < t; nr++){
        int n;
        cin >> n;

        if(n == 2) {
            cout << 1 << endl;
            cout << 2 << " " << 1 << endl;
            cout << 1 << endl;
            cout << 1 << " " << 2 << endl;
        } else if(n == 3){
            cout << 5 << endl;
            cout << 2 << " " << 3 << " " << 1 << endl;
            cout << 2 << endl;
            cout << 1 << " " << 3 << endl;
            cout << 2 << " " << 3 << endl;
        } else {
            int time = (n - 1) * (n - 1);
            for(int i = 0; i < n / 2 - 1; i++) time += 2 * (n - 2 - i) * (n - 2 - i);
            if(n % 2 == 1) time += (n / 2) * (n / 2);
            cout << time << endl;

            vector<int> p(n, 0);
            for(int i = 3; i <= n; i++) p[i - 2] = i;
            p[0] = n / 2 + 1;
            p[n / 2 - 1] = 1;
            p.back() = 2;

            for(int x : p) cout << x << " ";
            cout << endl;

            cout << n - 1 << endl;
            for(int i = 2; i <= n / 2; i++){
                cout << i << " " << n << endl;
            }
            for(int i = n / 2 + 1; i <= n - 1; i++){
                cout << i << " " << 1 << endl;
            }
            cout << 1 << " " << n << endl;
        }
    }
}