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 ll long long
#define pb push_back
#define mod 1000000007
#define F first
#define S second
#define all(v) (v).begin(),(v).end()
#define np next_permutation
#define lp(i,n) for(int i=0;i<n;i++)
#define lps(i,j,n) for(int i=j;i<n;i++)
#define vii vector<vi>
#define vb vector<bool>
#define pr pair<int,int>
#define vl vector<ll>
#define vs vector<string>
#define us unordered_map<int,int>
#define Mpq priority_queue<int>
#define mpq priority_queue<int,vi,greater<int>> 
#define eb emplace_back
#define pr pair<int,int>
#define prl pair<ll,ll>
#define vp vector<pr>
#define vpl vector<prl>
#define mkp make_pair
#define ld long double
#define vii vector<vi>
#define Max(a,b) a=max(a,b)
#define Min(a,b) a=min(a,b)
#define ull unsigned long long
#define prr pair<ll,int>

//#include <ext/pb_ds/tree_policy.hpp> 
//#include <ext/pb_ds/assoc_container.hpp>

//using namespace __gnu_pbds;
using namespace std;
using vi=vector<int>;

//typedef  tree<prl, null_type,less<prl>, rb_tree_tag, tree_order_statistics_node_update> ordered_set;
//ordered_set s;

#define F_OR(i, a, b, s) for (int i=(a); (s)>0?i<(b):i>(b); i+=(s))
#define F_OR1(e) F_OR(i, 0, e, 1)
#define F_OR2(i, e) F_OR(i, 0, e, 1)
#define F_OR3(i, b, e) F_OR(i, b, e, 1)
#define F_OR4(i, b, e, s) F_OR(i, b, e, s)
#define GET5(a, b, c, d, e, ...) e
#define F_ORC(...) GET5(__VA_ARGS__, F_OR4, F_OR3, F_OR2, F_OR1)
#define FOR(...) F_ORC(__VA_ARGS__)(__VA_ARGS__)
#define EACH(x, a) for (auto& x: a)
template<class T> using pqg = priority_queue<T,vector<T>,greater<T>>;

double epsilon=(double)1e-9;
const int N=(int)1e6+12;
int t,n,m;

const ll MX=(ll)1e18+123;
const int inf=0x3fffffff;
int a[N],b[N],cnt[N];
bool check(vl a){
   int n=a.size();
   lp(i,n-1){
      if(a[i]>a[i+1]) return 0;
      a[i+1]-=a[i];
      a[i]=0;
   }
   if(!a[n-1]) return 1;
   return 0;
}

void solve(){
   int n,j=1;
   ll ans=0;
   cin>>n;
   vi a(n+1);
   vp opr;
   lps(i,1,n+1) a[i]=i;

   for(int i=1;i<=(n+1)/2;i++){
      if(i==1){
	opr.eb(1,n);
	ans+=(n-1)*1LL*(n-1);
	swap(a[1],a[n]);
      }
      else{
	opr.eb(n-j,1);
	ans+=(ll)pow((ll)(n-j-1),2LL);
	swap(a[1],a[n-j]);
	if(i<(n+1)/2 or (i==(n+1)/2 and (n%2) == 0)){
           opr.eb(j+1,n);
           ans+=(ll)pow((ll)(n-j-1),2LL);
	   swap(a[j+1],a[n]);
	}
	j++;
      }
   }
   cout<<ans<<"\n";
   lps(i,1,n+1) cout<<a[i]<<" ";
   cout<<"\n";

   cout<<opr.size()<<"\n";
   for(int i=opr.size()-1;i>=0;i--){
      cout<<opr[i].F<<" "<<opr[i].S<<"\n";
   }

}

int main(){
   ios_base::sync_with_stdio(0);  
   cin.tie(nullptr);

   cin>>t;
   while(t--)
   solve(); 
   return 0;
}