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:
/**
 * Author: Daniel
 * Created Time: 2021-01-27 14:46:44
**/

// time-limit: 1000
#include <bits/stdc++.h>

using namespace std;

#define F first 
#define S second
#define ER erase
#define IS insert
#define PI acos(-1)
#define PB pop_back
#define EB emplace_back
#define lowbit(x) (x & -x)
#define SZ(x) ((int)x.size())
#define MP(x, y) make_pair(x, y)
#define ALL(x) x.begin(), x.end()
#define RALL(x) x.rbegin(), x.rend()
#define SOS; ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);cout<<fixed<<setprecision(10);

typedef long long LL;
typedef vector<LL> VL;
typedef vector<int> VI;
typedef pair<int, int> PII;
typedef unsigned long long ULL;

template <typename A> using VE = vector<A>;
template <typename A> using USET = unordered_set<A>;
template <typename A> using HEAP = priority_queue<A>;
template <typename A, typename B> using PA = pair<A, B>;
template <typename A, typename B> using UMAP = unordered_map<A, B>;
template <typename A> using RHEAP = priority_queue<A, vector<A>, greater<A> >;

///////////////////////////////////////////////////////////////////////////
//////////////////// DO NOT TOUCH BEFORE THIS LINE ////////////////////////
///////////////////////////////////////////////////////////////////////////

// check the limitation!!!
const int N = 100010, M = 1010;

int T;
int n;
int p[N];

// read the question carefully!!!
int main()
{
    SOS;

    cin >> T;
    while (T -- )
    {
        cin >> n;
        for (int i = 1; i <= n; i ++ ) p[i] = i;

        LL tot = 0;
        VE<PII> res;

        res.EB(1, n);
        swap(p[1], p[n]);
        tot += (1 - n) * (LL)(1 - n);

        int i = 2, j = n - 1;
        while (i <= j)
        {
            res.EB(i, n);
            swap(p[n], p[i]);
            tot += (n - i) * (LL)(n - i);
            if (i != j)
            {
                res.EB(j, 1);
                swap(p[1], p[j]);
                tot += (1 - j) * (LL)(1 - j);
            }
            i ++ , j -- ;
        }

        cout << tot << '\n';
        for (int i = 1; i <= n; i ++ )
        {
            cout << p[i];
            cout << " \n"[i == n];
        }
        cout << SZ(res) << '\n';
        reverse(ALL(res));
        for (auto [x, y] : res) cout << x << ' ' << y << '\n';
    }
    return 0;
}

// GOOD LUCK!!!