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 <bits/stdc++.h>
using namespace std;
#define INF 2000000000
typedef long long int ll;
typedef pair<ll,ll> ii;
typedef pair<ii,ll> iii;
typedef vector<ll> vi;
typedef vector<ii> vii;
#define getbit(n,i) (((n)&(1LL<<(i)))!=0)
#define setbit0(n,i) ((n)&(~(1LL<<(i))))
#define setbit1(n,i) ((n)|(1LL<<(i)))
#define lastone(n) ((n)&(-(n)))
#define read freopen("debug//in.txt","r",stdin)
#define write freopen("debug//out.txt","w",stdout)
#define DBG(a) cerr<<#a<<" ->->->-> "<<a<<"\n"
#define fi first
#define se second
#define PI (acos(-1))
#define fastread ios::sync_with_stdio(false);cin.tie(NULL)
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
#define mod (998244353)
#define N 1000005
template<class T,class V> ostream& operator<<(ostream &s,pair<T,V> a)
{
    s<<a.fi<<' '<<a.se;
    return s;
}
int main()
{
    fastread;
#ifdef FILE_IO2
    read;
    write;
#endif
    int T=1;
    cin>>T;
    for(int qq=1; qq<=T; qq++)
    {
        ll n;
        cin>>n;
        vi ans(n);
        ans[0] = n/2;
        ans.back() = n/2-1;
        for(int i=0;i<ans[0]-1;i++)ans[i+1] = i;
        for(int j=ans[0];j<n-1;j++)ans[j] = j+1;
        vii a;
        ll val = 0;
        for(int i=0;i<(n-1)/2;i++){
            a.emplace_back(ans[0]+1+i,1);
            val += (ans[0]+i)*(ans[0]+i);
        }
        for(int i=0;i<n/2-1;i++){
            a.emplace_back(ans.back()+1-i,n);
            val += (ans.back()+1-i-n)*(ans.back()+1-i-n);
        }
        a.emplace_back(1,n);
        val += (n-1)*(n-1);
        cout<<val<<endl;
        for(auto x:ans)cout<<x+1<<' ';
        cout<<endl;
        cout<<a.size()<<endl;
        for(auto x:a)cout<<x<<endl;
    }
}