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 FOR(i,a,b) for(int i=a;i<=b;i++)
#define ROF(i,a,b) for(int i=a;i>=b;i--)
#define ll         long long
#define D          double
#define P          pair<int,int>
#define Pll        pair<long long,long long>
#define fi         first
#define se         second
//-----------------------------------------math----------------------------------------
#define MAX(x,y)   x=max(x,y)
#define MIN(x,y)   x=min(x,y)
#define pow2(x)    (1<<(x))
//-----------------------------------------STL-------------------------------------------
#define pb         push_back
#define ALL(a)     a.begin(),a.end()
#define POS(a,c)   lower_bound( a.begin() , a.end() , (c) ) - a.begin() + 1
#define UNI(a)     sort(ALL(a)),a.erase(unique(ALL(a)),a.end())
#define DELF(a)    a[0]=a[SZ(a)-1],a.resize(SZ(a)-1)
#define SUM(a,l,r) upper_bound(ALL(a),r)-lower_bound(ALL(a),l)
#define E(a,b)     a.find(b)!=a.end()
//---------------------------------------------io-------------------------------------------
#define read(x)    scanf("%d",&x)
#define FLUSH      fflush(stdout)
#define TC         int T;cin>>T;while(T--)
#define watch(x)   cout<<(#x)<<"  is  "<<x<<endl
#define END        {cout<<-1;return 0;}
//---------------------------------------segment tree---------------------------------------------
#define callmid    int mid=(L+R)>>1
#define ls         p+p
#define rs         p+p+1
#define rmid       r<=mid
#define midl       l>=mid+1
#define xmid       x<=mid
#define lsg        L,mid
#define rsg        mid+1,R
//-----------------------------------------graph---------------------------------------------------
#define read_edge  read(a),read(b),g[a].pb(b),g[b].pb(a)
#define NOT(u)     if(v==u)continue
#define notfa      if(v==fa)continue
//------------------------------------------else---------------------------------------
#define all(a)     a+1,a+n+1
#define SZ(x)      (int)(x).size()
#define mod        998244353
#define modd       1000000007
using namespace std;

P operator*(const P&p,const P&x){P np=make_pair(p.fi*x.fi,p.se*x.se);return np;}
Pll operator%(const Pll&p,const Pll&x){Pll np=make_pair(p.fi%x.fi,p.se%x.se);return np;}
Pll operator*(const Pll&p,const P&x){Pll np=make_pair(p.fi*x.fi,p.se*x.se);return np;}
Pll operator*(const Pll&p,const Pll&x){Pll np=make_pair(p.fi*x.fi,p.se*x.se);return np;}
Pll operator+(const Pll&p,const Pll&x){Pll np=make_pair(p.fi+x.fi,p.se+x.se);return np;}
Pll operator-(const Pll&p,const Pll&x){Pll np=make_pair(p.fi-x.fi,p.se-x.se);return np;}
P operator+(const P&p,const P&x){P np=make_pair(p.fi+x.fi,p.se+x.se);return np;}
P operator-(const P&p,const P&x){P np=make_pair(p.fi-x.fi,p.se-x.se);return np;}

vector<int> prime;
bool sortinrev(const pair<int,int> &a,const pair<int,int> &b){return (a.first > b.first);}
bool sortbysecdesc(const pair<int,int> &a,const pair<int,int> &b){return a.second>b.second;}
bool sortbysec(const pair<int,int> &a,const pair<int,int> &b) {return (a.second < b.second);}
ll powerm(ll a, ll k) { if (!k) {return 1;}ll b = powerm(a, k / 2);b = b * b % mod;if (k % 2) {return a * b % mod;} else {return b;}}
ll power(ll a , ll b) { if(b == 1) return a; if(b == 0) return 1; ll m1 = power(a,b/2); if(b%2) return m1*m1*a; return m1*m1; }
void PFactor(ll n){prime.clear();FOR(i,2,n){while(n!=i){if(n%i)break;else prime.pb(i),n/=i;}}prime.pb(n);}
bool isprime (ll a) { if(a<=1) return false; if(a==2||a==3) return true; if(a%2==0||a%3==0) return false; for(ll i=5;i*i<=a;i=i+6) { if(a%i==0||a%(i+2)==0) return false; } return true;}




const int N = 2e5 + 20;
vector<int> g[N];
int n=0,m=0;
int a;
int b;
P ans[N];
int p[N];
int main()
{
    TC
    {
        cin>>n;
        FOR(i,1,n)p[i]=i;
        int cnt=0;
        ll sum=0;
        ans[++cnt]=P(1,n);
        swap(p[1],p[n]);

        FOR(i,2,n/2)
        {
            ans[++cnt]=P(i,n);
            swap(p[i],p[n]);
            ans[++cnt]=P(n-i+1,1);
            swap(p[1],p[n-i+1]);
        }

        if(n%2)swap(p[n/2+1],p[n]),ans[++cnt]=P(n/2+1,n);
        FOR(i,1,cnt)sum+=power(ans[i].fi-ans[i].se,2);
        cout<<sum<<endl;
        FOR(i,1,n)cout<<p[i]<<" ";
        cout<<"\n"<<cnt<<endl;
        ROF(i,cnt,1)
        {
            cout<<ans[i].fi<<' '<<ans[i].se<<endl;
        }
    }
}