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;
 
typedef long long ll;
typedef vector<int> vi;
typedef vector<pair<int, int>> vii;
typedef pair<int, int> pi;
#define MOD 1000000007LL

int t, n;

int main() {
    //freopen("feast.in", "r", stdin);
    //freopen("feast.out", "w", stdout);
    //ios_base::sync_with_stdio(false);
    //cin.tie(0);
    
    cin >> t;
    while (t--) {
        cin >> n;
        vector<pair<int, int>> edges;
        vector<int> p(n+1);
        ll ans = 0;
        for (int i = 1; i <= n; i++) p[i] = i;
        for (int i = (n+2)/2; i <= n; i++) {
            edges.push_back({i, 1});
            swap(p[i], p[1]);
            ll dis = i-1;
            ans += dis*dis;
        }
        for (int i = 2; i < (n+2)/2; i++) {
            edges.push_back({i, n});
            swap(p[i], p[n]);
            ll dis = n-i;
            ans += dis*dis;
        }
        cout << ans << endl;
        for (int i = 1; i < n; i++) cout << p[i] << " ";
        cout << p[n] << endl;
        cout << n-1 << endl;
        for (int i = n-2; i >= 0; i--) cout << edges[i].first << " " << edges[i].second << endl;
        
    }
}