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 all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define vi vector<int>
#define sz(x) (int)(x).size()
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
mt19937 rnd(chrono::high_resolution_clock::now().time_since_epoch().count());
template<class T> bool cmn(T& a, T b) { return b < a ? a = b, true : false; }
template<class T> bool cmx(T& a, T b) { return b > a ? a = b, true : false; }
void solve() {
    int n;
    cin >> n;
    int mid = n / 2 - 1;
    if (n == 2) {
        cout << "1\n2 1\n1\n1 2\n";
        return;
    }
    if (n == 3) {
        cout << "5\n2 3 1\n2\n1 3\n2 3\n";
        return;
    }
    vector<int> p(n);
    for (int i = 1; i < mid; i++) p[i] = i + 2;
    p[mid] = 1;
    p[n - 1] = 2;
    p[0] = mid + 2;
    for (int i = mid + 1; i < n - 1; i++) p[i] = i + 2;
    vector<pii> ans;
    for (int i = 1; i <= mid; i++) ans.pb(mp(i + 1, n));
    for (int i = mid + 1; i < n; i++) ans.pb(mp(i + 1, 1));
    ll S = 0;
    for (auto [x, y] : ans) S += 1ll * (x - y) * (x - y);
    cout << S << '\n';
    for (int i : p) cout << i << ' ';
    cout << '\n';
    cout << sz(ans) << '\n';
    for (auto [x, y] : ans) cout << x << ' ' << y << '\n';
}
int main() {
    ios::sync_with_stdio(0);
    cin.tie(0), cout.tie(0);
    int t;
    cin >> t;
    while (t--) solve();
    return 0;
}