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 <cstdio>
#include <cmath>
#include <ctime>
#include <cstdlib>
#include <cassert>
#include <vector>
#include <list>
#include <stack>
#include <queue>
#include <deque>
#include <map>
#include <set>
#include <bitset>
#include <string>
#include <algorithm>
#include <utility>
#include <complex>
#define rep(x, s, t) for(llint (x) = (s); (x) <= (t); (x)++)
#define chmin(x, y) (x) = min((x), (y))
#define chmax(x, y) (x) = max((x), (y))
#define all(x) (x).begin(),(x).end()
#define inf 1e18

using namespace std;

typedef long long llint;
typedef long long ll;
typedef pair<llint, llint> P;

ll T;
ll n;
ll p[200005];
vector<P> ans;

ll add(ll x, ll y)
{
	ans.push_back(P(y, x));
	swap(p[x], p[y]);
	return (x-y)*(x-y);
}

int main(void)
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	
	cin >> T;
	while(T--){
		cin >> n;
		rep(i, 1, n) p[i] = i;
		
		if(n == 1){
			cout << 0 << endl;
			cout << 1 << endl;
			cout << 0 << endl;
			continue;
		}
		
		ll cost = 0; ans.clear();
		cost += add(1, n);
		rep(i, n/2+1, n-1) cost += add(1, i);
		rep(i, 2, n/2) cost += add(n, i);
		reverse(all(ans));
		
		cout << cost << "\n";
		rep(i, 1, n) cout << p[i] << " "; cout << "\n";
		cout << ans.size() << "\n";
		for(auto p : ans) cout << p.first << " " << p.second << "\n";
		flush(cout);
		
		for(auto q : ans){
			ll i = q.first, j = q.second;
			assert(i != j && p[j] == i);
			swap(p[i], p[j]);
		}
		rep(i, 1, n) assert(p[i] == i);
	}
	flush(cout);
	
	return 0;
}