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 <iostream>
#include <vector>
#include <cmath>
#include <algorithm>
#include <set>
#include <map>

#define ll long long
#define ld double
#define ft first
#define sd second
#define pb push_back
#define eb emplace_back

//#pragma GCC optimize("Ofast")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,tune=native")
//#pragma GCC optimize("unroll-loops")
//#pragma GCC optimize("fast-math")

using namespace std;

int main() {
    ios::sync_with_stdio(0);
    cin.tie(NULL);
    cout.tie(NULL);
#ifdef FREOPEN
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
#endif

    ll t;

    cin >> t;

    for (ll step = 0; step < t; ++step) {
        ll n;

        cin >> n;
        vector <ll> p;
        vector <pair<ll, ll>> op;
        for (ll i = 1; i <= n; ++i)
            p.pb(i);

        ll l = 0;
        ll r = n-1;
        ll ans = 0;
        while (l <= r) {
            if (r <= n-1-l) {
                swap(p[l], p[n-1]);
                ans += (n-1-l)*(n-1-l);
                op.pb({l+1,n});
                ++l;
                if (r == n-1)
                    --r;
            }
            else {
                swap(p[r],p[0]);
                ans += r*r;
                op.pb({r+1, 1});
                --r;
            }
        }
        reverse(op.begin(), op.end());
        cout << ans << '\n';
        for (auto &i: p) {
            cout << i << ' ';
        }
        cout << '\n' << op.size() << '\n';
        for (auto &i: op) {
            cout << i.ft << ' ' << i.sd << '\n';
        }
    }
    return 0;
}