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>

#define int long long

using namespace std;

#ifdef LOCAL 
#include "/media/brayand/BrayanD/debugger.h"
#else
#define db(...) false
#define dbl(...) false
#define dbg(...) false
#define dbm(...) false
#define dbs(...) false
#define dbas(...) false
#endif

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);

    int T;
    cin >> T;

    while(T--)
    {
    	int N;
    	cin >> N;

        if(N == 2)
        {
            cout << "1" << '\n';
            cout << "2 1" << '\n';
            cout << "1" << '\n';
            cout << "2 1" << '\n';
            continue;
        }
        if(N == 3)
        {
            cout << "5" << '\n';
            cout << "2 3 1" << '\n';
            cout << "2" << '\n';
            cout << "1 3" << '\n';
            cout << "3 2" << '\n';
            continue;
        }

    	vector<int> vect;

        int sz1 = (N-2)/2;
        int sz2 = (N-2)-sz1;

        vect.push_back(1+sz1+1);

        for(int i = 3 ; i <= 1+sz1 ; i++)
        {
            vect.push_back(i);
        }

        vect.push_back(1);

        for(int i = 1+sz1+2 ; i <= N ; i++)
        {
            vect.push_back(i);
        }

        vect.push_back(2);

        int sum = 0;

        for(int i = 1, ppp = N-2 ; i <= sz1 ; i++, ppp--)
        {
            sum += ppp*ppp;
        }  

        for(int i = 1, ppp = N-2 ; i <= sz2 ; i++, ppp--)
        {
            sum += ppp*ppp;
        }  

        sum += (N-1)*(N-1);

        cout << sum << '\n';

        for(auto x : vect)
        {
            cout << x << " ";
        }
        cout << '\n';

        cout << N-1 << '\n';

        for(int i = 1 ; i <= sz1 ; i++)
        {
            cout << i+1 << " " << N << '\n';
        }

        for(int i = 1 ; i <= sz2 ; i++)
        {
            cout << sz1+1+i << " " << 1 << '\n';
        }

        cout << 1 << " " << N << '\n';
    }

    return 0;
}