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:
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <algorithm>
#include <cmath>
#include <iomanip>
#include <string>
#include <fstream>
#include <vector>
#include <queue>
#include <stack>
#include <sstream>
#include <list>
#include <map>
#include <set>
#include <limits>
#include <random>
#include <functional>
#include <unordered_set>
#include <unordered_map>

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef unsigned int ui;

#define pii pair<int, int>
#define pll pair<ll, ll>
#define vi vector<int>
#define vl vector<ll>

using namespace std;

const int N = 1000010;
const int MOD = 1000000007;
const int INF = 1009000999;
const ll LINF = (1ll << 60) + 1337;
const ld EPS = 0.0000000001;

int main() {
    ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#ifdef _DEBUG
    freopen("input.txt", "r", stdin); freopen("output.txt", "w", stdout);
#endif
    int tt;
    cin >> tt;
    while (tt--) {
        int n;
        cin >> n;
        ll cost = 0;
        vector<pii> ans;
        vector<int> p(n + 1);
        for (int i = 0; i <= n; i++) p[i] = i;
        for (int i = 1; i <= n; i++) {
            if (i == p[i]) {
                if (i - 1 > n - i) {
                    ans.push_back({ i,1 });
                    swap(p[i], p[1]);
                    cost += (i - 1) * 1ll * (i - 1);
                }
                else {
                    ans.push_back({ i, n });
                    swap(p[n], p[i]);
                    cost += (n - i) * 1ll * (n - i);
                }
            }
        }
        cout << cost << "\n";
        for (int i = 1; i <= n; i++) {
            cout << p[i] << " ";
        }
        cout << "\n" << ans.size() << "\n";
        reverse(ans.begin(), ans.end());
        for (auto& i : ans) cout << i.first << " " << i.second << "\n";
    }
}

/*
    Important stuff
    - int overflow
    - array bounds
    - special cases (n=1, max n) maybe adhoc problem
    - doubles are read for a long time (4 * 10 ^ 5 danger GNU vs MS ICPC 2020 1/4)
    - don't get stuck on one approach
    - don't get stuck on one problem
    - recheck submissions if near the end of the tournament and there is nothing to do
    - do something instead of nothing and stay organized
*/