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 <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstring>
#include <deque>
#include <fstream>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>

//#define ENVIRONMENT_LINKED_ACL

#ifdef ENVIRONMENT_LINKED_ACL
#include <atcoder/convolution>
#include <atcoder/lazysegtree>
#include <atcoder/segtree.hpp>
#endif

//#define ENVIRONMENT_LINKED_BOOST

#ifdef ENVIRONMENT_LINKED_BOOST
#include <boost/multiprecision/cpp_dec_float.hpp>
#include <boost/multiprecision/cpp_int.hpp>
#endif

using namespace std;
using uint = unsigned int;
using ll   = long long;
using ull  = unsigned long long;
#define REP(i, a, b)    for (ll i = a; i < b; ++i)
#define REPREV(i, a, b) for (ll i = a; i > b; --i)

const int _ = []() {
    std::cin.tie(nullptr);
    std::ios::sync_with_stdio(false);
    std::cout << std::fixed << std::setprecision(10);
    return 0;
}();

template <typename value_t>
void resize(value_t& v, const value_t& val) { v = val; }
template <typename vec_t, typename value_t, typename... arg_t>
void resize(std::vector<vec_t>& v, const value_t& val, int size, arg_t... arg) {
    v.resize(size);
    for (auto& c : v) resize(c, val, arg...);
}

template <typename A, typename B>
void chmin(A& a, const B& b) {
    a = min(a, static_cast<A>(b));
};
template <typename A, typename B>
void chmax(A& a, const B& b) {
    a = max(a, static_cast<A>(b));
};

int main() {
    ll t;
    cin >> t;
    while (t--) {
        ll n;
        cin >> n;
        if (n == 2) {
            cout << "1\n2 1\n1\n2 1" << endl;
            continue;
        }
        if (n == 3) {
            cout << "5\n2 3 1\n2\n1 3\n2 3" << endl;
            continue;
        }

        ll l   = n / 2;
        ll r   = n / 2 + 1;
        ll ans = 0;
        vector<pair<int, int>> op;
        REP(i, r, n) {
            ans += (i - 1) * (i - 1);
            op.push_back({i, 1});
        }
        REPREV(i, l, 0) {
            ans += (n - i) * (n - i);
            op.push_back({i, n});
        }
        cout << ans << endl;
        cout << r << " ";
        REP(i, 1, n + 1) {
            if (i == l || i == r) continue;
            cout << i << " ";
        }
        cout << l << endl;
        cout << op.size() << endl;
        for (auto p : op) {
            cout << p.first << " " << p.second << endl;
        }
    }

    return 0;
}