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>
using namespace std;

typedef long long int ll;
typedef long double ld;
typedef pair<ll, ll> pl;
typedef pair<ld, ld> pd;
typedef priority_queue<pl, vector<pl>, greater<pl> > pq;

template<class T, class S>
ostream& operator<<(ostream &o, pair<T, S> p) {
    return o<<'('<<p.first<<", "<<p.second<<')';
}

template<template<class, class...> class T, class... A>
typename enable_if<!is_same<T<A...>, string>(), ostream&>::type
operator<<(ostream &o, T<A...> V) {
	o<<'[';
	for(auto a:V) o<<a<<", ";
	return o<<']';
}

// This is tuples, _p is helper
template<ll i, class... T>
typename enable_if<i==sizeof...(T)>::type _p(ostream& o, tuple<T...> t) {}

template<ll i, class... T>
typename enable_if<i<sizeof...(T)>::type _p(ostream& o, tuple<T...> t) {
    _p<i+1>(o << get<i>(t)<< ", ", t);
}
 
template<class... T>
ostream& operator<<(ostream& o, tuple<T...> t) {
    _p<0>(o<<'(', t);
    return o<<')';
}


#define G(x) ll x; cin >> x;
#define Gs(s) string s; cin >> s;
#define Gd(x) ld x; cin >> x;
#define F(i, l, r) for(ll i = l; i < (r); i++)
#define F1(i, l, r) for(ll i = l; i >= (r); i--)
#define Fi(i, s) for(auto i = (s).begin(); i != (s).end(); i++)
#define L G(_t) F(_t1, 0, _t)
#define H(ar, n) vector<ll> ar(n); F(_i, 0, n) cin >> ar[_i];
#define A first
#define B second
#define PA(ar, n) F(_i, 0, n) cout << (ar)[_i] << " "; cout << endl; 
#define PB push_back
#define V vector
#define CASE cout << "Case #" << _t1 + 1 << ": " 
#define CC << " " <<
#define BE(ar) ar.begin(), ar.end()
#define CO(x) cout << (x) << endl;
#define D(x) ((x) - '0')
#define EX(x) {cout << (x) << endl; return 0;}
#define S(ar, val) memset(ar, val, sizeof(ar));
ll M = 1000000007; //998244353;
ll inv(ll a, ll b){return 1<a ? b - inv(b%a,a)*b/a : 1;}
ll gcd(ll x, ll y){if(y == 0) return x; return gcd(y, x % y); }
pl neigh[4] = {{-1, 0}, {1, 0}, {0, 1}, {0, -1}};

#define N 0

int main() {
        ios_base::sync_with_stdio(0);
        cin.tie(0);
	L{
		G(n)
		if(n == 2){
			cout << "1\n2 1\n1\n2 1" << endl;
			continue;
		}
		if(n == 3){
			cout << "5\n2 3 1\n2\n1 3\n3 2" << endl;
			continue;
		}
		ll perm[n];
		perm[n-1] = 1;
		perm[0] = n-1;
		F1(i, n-2, (n-1)/2+2){
			perm[i] = i;
		}
		perm[(n-1)/2] = n;
		perm[(n-1)/2+1]=2;
		F(i, 1, (n-1)/2){
			perm[i] = i+2;
		}
		ll x = (n-1)*(n-1);
		ll c = 1;
		ll p = n-1;
		while(c < n-1){
			if(c %2) p--;
			c++;
			x += p * p;
		}
		cout << x << endl;
		F(i, 0, n){
			cout << perm[i] << " ";
		}
		cout << endl;
		cout << n - 1 << endl;
		F1(i, n-1, (n-1)/2+2){
			cout << i << " " << 1 << endl;
		}
		F(i, 1, (n-1)/2+2){
			cout << i << " " << n << endl;
		}
	}
}