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:
/*
#pragma GCC target("avx2")
#pragma GCC optimize("O3")
#pragma GCC optimize("unroll-loops")
//*/
 
#include <bits/stdc++.h>
 
// #include <atcoder/all>
 
using namespace std;
// using namespace atcoder;
 
#define DEBUG(x) cerr<<#x<<": "<<x<<endl;
#define DEBUG_VEC(v) cerr<<#v<<":";for(int i=0;i<v.size();i++) cerr<<" "<<v[i]; cerr<<endl;
#define DEBUG_MAT(v) cerr<<#v<<endl;for(int i=0;i<v.size();i++){for(int j=0;j<v[i].size();j++) {cerr<<v[i][j]<<" ";}cerr<<endl;}
typedef long long ll;
#define int ll
 
#define vi vector<int>
#define vl vector<ll>
#define vii vector< vector<int> >
#define vll vector< vector<ll> >
#define vs vector<string>
#define pii pair<int,int>
#define pis pair<int,string>
#define psi pair<string,int>
#define pll pair<ll,ll>
template<class S, class T> pair<S, T> operator+(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first + t.first, s.second + t.second); }
template<class S, class T> pair<S, T> operator-(const pair<S, T> &s, const pair<S, T> &t) { return pair<S, T>(s.first - t.first, s.second - t.second); }
template<class S, class T> ostream& operator<<(ostream& os, pair<S, T> p) { os << "(" << p.first << ", " << p.second << ")"; return os; }
#define X first
#define Y second
#define rep(i,n) for(int i=0;i<(int)(n);i++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)
#define rrep(i,n) for(int i=(int)(n)-1;i>=0;i--)
#define rrep1(i,n) for(int i=(int)(n);i>0;i--)
#define REP(i,a,b) for(int i=a;i<b;i++)
#define in(x, a, b) (a <= x && x < b)
#define all(c) c.begin(),c.end()
void YES(bool t=true) {cout<<(t?"YES":"NO")<<endl;}
void Yes(bool t=true) {cout<<(t?"Yes":"No")<<endl;}
void yes(bool t=true) {cout<<(t?"yes":"no")<<endl;}
void NO(bool t=true) {cout<<(t?"NO":"YES")<<endl;}
void No(bool t=true) {cout<<(t?"No":"Yes")<<endl;}
void no(bool t=true) {cout<<(t?"no":"yes")<<endl;}
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a = b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a = b; return 1; } return 0; }
#define UNIQUE(v) v.erase(std::unique(v.begin(), v.end()), v.end());
const ll inf = 1000000001;
const ll INF = (ll)1e18 + 1;
const long double pi = 3.1415926535897932384626433832795028841971L;
int popcount(ll t) { return __builtin_popcountll(t); }
// int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
// int dx2[8] = { 1,1,0,-1,-1,-1,0,1 }, dy2[8] = { 0,1,1,1,0,-1,-1,-1 };
vi dx = {0, 1, 0, -1}, dy = {-1, 0, 1, 0};
// vi dx2 = { 1,1,0,-1,-1,-1,0,1 }, dy2 = { 0,1,1,1,0,-1,-1,-1 };
struct Setup_io {
    Setup_io() {
        ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
        cout << fixed << setprecision(25);
    }
} setup_io;
const ll MOD = 1000000007;
// const ll MOD = 998244353;
// #define mp make_pair
//#define endl '\n'


void solve() {
    int n;
    cin >> n;
    vi a(n);
    rep (i, n) a[i] = i;

    vector<pii> op;
    if (n % 2 == 1) {
        int cen = n / 2;
        op.push_back(pii(cen, 0));
        for (int diff = 1; cen - diff > 0; diff++) {
            op.push_back(pii(cen - diff, n - 1));
            op.push_back(pii(cen + diff, 0));
        }
        op.push_back(pii(0, n - 1));
    }
    else {
        for (int diff = 1; n / 2 - diff > 0; diff++) {
            op.push_back(pii(n / 2 - diff, n - 1));
            op.push_back(pii(n / 2 + diff - 1, 0));
        }
        op.push_back(pii(0, n - 1));
    }
    // DEBUG_VEC(op);

    ll ans = 0;
    rrep (i, op.size()) {
        swap(a[op[i].first], a[op[i].second]);
        ans += (op[i].second - op[i].first) * (op[i].second - op[i].first);
    }
    // DEBUG_VEC(a);

    cout << ans << '\n';
    rep (i, n) cout << a[i] + 1 << " ";
    cout << '\n';
    cout << op.size() << '\n';
    rep (i, op.size()) {
        cout << op[i].first + 1 << " " << op[i].second + 1 << '\n';
        assert(a[op[i].second] == op[i].first);
        swap(a[op[i].first], a[op[i].second]);
    }
    // DEBUG_VEC(a);
}

signed main() {
    int t = 1;
    cin >> t;
    
    while (t--) solve();
}