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>

typedef long long ll;
typedef long double ld;
using namespace std;
mt19937_64 mrand(chrono::steady_clock::now().time_since_epoch().count());
#define ii for(int i=1;i<=n;++i)
#define ji for(int j=1;j<=n;++j)
#define jj for(int j=1;j<=m;++j)
#define ij for(int i=1;i<=m;++i)
#define al(x) x.begin(),x.end()
#define asd cout<<"ok"<<endl;
#define vi vector<int>
#define vvi vector<vector<int>>
#define vl vector<ll>
#define vii vector<pair<int,int>>
#define pc(x) __builtin_popcount(x)
#define pb push_back

int main() {
    ios::sync_with_stdio(0);cin.tie(0);cout.tie(0);
    int t;
    cin >> t;
    while(t--) {
        int n;
        cin >> n;
        vector<int> mid((n-1)/2);
        iota(al(mid), n-mid.size());
        vector<int> pre(n-mid.size()-2);
        iota(al(pre), 0);
        vector<int> a;
        a.push_back(n-mid.size()-1);
        for(int i:pre) a.push_back(i);
        for(int i:mid) a.push_back(i);
        a.push_back(n-mid.size()-2);
        vector<int> aa = a;
        ll ans = 0;
        vector<pair<int,int>> b;
        for(int i=0;i<(n-1)/2;++i) {
            ans += 1ll * a[0] * a[0];
            int tmp = a[0];
            swap(a[0], a[tmp]);
            b.push_back({tmp, 0});
        }
        for(int i=0;i<n-mid.size()-2;++i) {
            ans += 1ll * (n-1-a[n-1]) * (n-1-a[n-1]);
            int tmp = a[n-1];
            swap(a[n-1], a[tmp]);
            b.push_back({tmp, n-1});
        }
        ans += 1ll * (n-1) * (n-1);
        swap(a[0], a[n-1]);
        b.push_back({0, n-1});
        assert(is_sorted(al(a)));
        cout << ans << endl;
        for(int i : aa) cout << i + 1 << " "; cout << endl;
        cout << b.size() << endl;
        for(auto &p : b) cout << p.first + 1 << " " << p.second + 1 << '\n';
    }
}