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<cstring>
//#include<vector>
//#include<queue>
//#include<sstream>
//#include<cmath>
//#include<set>
//#include<cstdio>
//#include<algorithm>
//#include<map>
#include<bits/stdc++.h>
#define mp make_pair
#define pii pair<int,int>
#pragma GCC optimize ("O2")
#define pdd pair<double,double>
#define pll pair<ll,ll>
#define mst(a,b) memset(a,b,sizeof(a))
#pragma comment(linker, "/STACK:102400000,102400000")
//g++ strlen.cpp -o a.exe -Wl,--stack,536870912
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<double> vec;
typedef vector<vec> mat;
const double eps = 1e-8;
const int maxn = 1e5+9;
const double pi=acos(-1.0);
const int inf = 5e8;
const ll mod = 1e9+7;
int op[maxn];
int main()
{
    ios::sync_with_stdio(false);
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        if(n==2){
            cout<<"1\n2 1\n1\n1 2\n";
            continue;
        }
        else if(n==3){
            cout<<"5\n2 3 1\n2\n1 3\n3 2\n";
            continue;
        }
        else if(n==4){
            cout<<"17\n3 1 4 2\n3\n3 1\n2 4\n1 4\n";
            continue;
        }
        int tmp=n-2;
        ll ans=1ll*(n-1)*(n-1);
        for(int i=n-2;i>=0;--i){
            ans+=1ll*i*i;
            --tmp;
            if(!tmp) break;
            ans+=1ll*i*i;
            --tmp;
            if(!tmp) break;
        }
        cout<<ans<<"\n";
        if(n&1) tmp=(n+1)/2;
        else tmp=n/2+1;
        op[1]=tmp;
        for(int i=tmp;i<n-1;++i) op[i]=i+1;
        op[n-1]=2;
        op[n]=1;
        for(int i=2;i<tmp-1;++i) op[i]=i+1;
        op[tmp-1]=n;
        for(int i=1;i<=n;++i) cout<<op[i]<<" ";
        cout<<"\n";
        cout<<n-1<<"\n";
        for(int i=tmp;i<n;++i) cout<<i<<" 1\n";
        for(int i=1;i<tmp;++i) cout<<i<<" "<<n<<"\n";
    }
    return 0;
}