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:
#define IO ios::sync_with_stdio(false);cin.tie();cout.tie(0)
#pragma GCC optimize(2)
#include<set>
#include<map>
#include<cmath>
#include<stack>
#include<queue>
#include<bitset>
#include<random>
#include<bitset>
#include<string>
#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<unordered_map>
#include<unordered_set>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll,int> pli;
typedef pair<int,int> pii;
const ll mod=1e9+7;
// int prime[1000010],cnt;
// bool st[1000010];
// void init(int n)
// {
//     for(int i=2;i<=n;i++)
//     {
//         if(!st[i]) prime[++cnt]=i;
//         for(int j=1;prime[j]<=n/i;j++)
//         {
//             st[prime[j]*i]=1;
//             if(i%prime[j]==0) break;
//         }
//     }
// }
int p[200010];
int main()
{
    IO;
    int T=1;
    cin>>T;
    //init(1000000);
    while(T--)
    {
        int n;
        cin>>n;
        for(int i=1;i<=n;i++) p[i]=i;
        vector<pii> ans;
        ll cost=1ll*(n-1)*(n-1);
        ans.push_back({1,n});
        swap(p[1],p[n]);
        
        for(int i=2;i<=n/2;i++) 
        {
            cost+=1ll*(n-i)*(n-i);
            ans.push_back({i,n});
            swap(p[i],p[n]);
        }
        for(int i=n/2+1;i<n;i++) 
        {
            cost+=1ll*(i-1)*(i-1);
            ans.push_back({i,1});
            swap(p[i],p[1]);
        }
        reverse(ans.begin(),ans.end());
        cout<<cost<<'\n';
        for(int i=1;i<=n;i++) cout<<p[i]<<' ';
        cout<<'\n'<<ans.size()<<'\n';
        for(auto t:ans) cout<<t.first<<' '<<t.second<<'\n';
    }
    return 0;
}