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;
#pragma GCC optimize("Ofast")
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
#define FASTER ios_base::sync_with_stdio(0);cin.tie(0)
#define setPrecision(n) cout << std::setprecision(n) << std::fixed;
#define rep(i,j,n) for(int i = j; i < n; i++)
#define ll long long
#define pb push_back
#define ppb pop_back
#define pii pair<int,int>
#define pli pair<long long,int>
#define vi vector<int>
#define vll vector<ll>
#define sz(x) ((int) (x).size())
#define all(c) (c).begin(),(c).end()
#define F first
#define S second
#define INF 2e9
#define LINF 2e18
const long long MOD = 1e9 + 7;
const double err = 1e-10;
const int N = 1e5 + 5;

//bruh what is pj  = i. Lame Lame Lame Lame

void solve() {
    int n; cin >> n;
    ll ans = 0;

    //maximum n-1 such operations
    // (n-1)^2 -> only one such operation
    // (n-2)^2 ->

    vi V(n);
    iota(all(V), 1);
    vector<pii> ways;
    vector<bool> swapped(n, 0);
    ways.pb({0, n-1});
    ans += 1ll * (n-1) * (n-1);
    swap(V[0], V[n-1]);
    swapped[0] = swapped[n-1] = 1;
    int l = 1, r = n-2;
    while(l <= r){
        if(!swapped[l]){
            swap(V[l], V[n-1]);
            ways.pb({n-1, l});
            ans += 1ll * (n-1 - l) * (n-1 - l);
            swapped[l] = 1;
        }
        if(!swapped[r]){
            swap(V[0], V[r]);
            ways.pb({0, r});
            ans += 1ll * (r) * (r);
            swapped[r] = 1;
        }
        l++, r--;
    }
    cout << ans << endl;
    rep(i,0,n) cout << V[i] << ' '; cout << endl;
    cout << sz(ways) << endl;
    reverse(all(ways));
    for(auto P : ways){
        cout << P.S + 1 << ' ' << P.F + 1 <<endl;
    }
}

int main() {
    // freopen("input.txt", "r", stdin);
    // freopen("output.txt", "w", stdout);
    FASTER;
    int T = 1;
    cin >> T;
    while (T--) solve();
    return 0;
}