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>
#define watch(x) cout <<(#x)<<" is "<<(x)<<endl
#define debug cout <<"hi"<<endl
#define maxn 3e5+10
#define inf32 0x3f3f3f3f

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<ll, ll> pll;
typedef pair<int,int>pii;
const int MOD=1e9+7;
const int INF32=1<<30;
const ll INF64=1LL<<60;

void solve(){
    int tc;cin >> tc;
    while(tc--){
        int n;cin >> n;
        vector<int>p(n,0);
        if(n%2){p[n/2] = 1;p[n/2+1] = n;}
        else {p[n/2-1] = 1;p[n/2] = n;}
        p[0] = n-1;p[n-1] = 2;
        int now = 1,cnt = 3;
        while(p[now]==0){p[now] = cnt;now++;cnt++;}
        now = n-2;cnt = n-2;
        while(p[now]==0){p[now] = cnt;now--;cnt--;}
        if(n==2){p[0] = 2;p[1] = 1;}
        else if(n==3){p[0] = 2;p[1] = 3;p[2] = 1;}
        ll ans = 0;cnt = n-2;
        ans += 1ll*(n-1)*1ll*(n-1);
        for(ll i = n-2;cnt>0;i--){
            if(cnt==1) ans += i*i;
            else ans += 2*i*i;
            cnt -= 2;
        }
        cout << ans <<endl;
        for(int i = 0;i<n;i++) cout << p[i] <<' ';
        cout <<'\n';
        cout << n-1 <<'\n';
        if(n==3){
            cout << 2 <<' '<< 1 <<endl;
            cout << 1 <<' '<< 3 <<endl;continue;
        }
        for(int i = 2;i<=n/2+n%2;i++) cout << i <<' '<< n <<'\n';
        for(int i = n-1;i>n/2+n%2;i--) cout << i <<' '<< 1 <<'\n';
        cout << 1 <<' '<< n <<'\n';
    }
}

int main(){
  ios_base::sync_with_stdio(0);
  cin.tie(0); cout.tie(0);
  solve();
  return 0;
}