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>
using namespace std;
#define endl "\n"
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int TT;scanf("%d",&TT);for(int TTT=1;TTT<=TT;TTT++)
#define fastio()  ios::sync_with_stdio(false),cin.tie(0),cout.tie(0)
#define forrr(i,x,y) for(int i=x;i<=y;++i)
#define rofff(i,x,y) for(int i=x;i>=y;--i)
#define forr(i,x,y) for(int i=x;i<y;++i)
#define roff(i,x,y) for(int i=x;i>y;--i)
#define lowbit(x) (x&-x)
#define fin(FI) freopen(FI"r",stdin)
#define fout(FI) freopen(FI,"w",stdout)
#define fi first
#define se second
#define mp make_pair
#define sf scanf
#define pf printf
#define pb push_back
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
typedef pair<string,int> psi;
const ldb pi = acos((ldb)-1);
void in(initializer_list<int*> li) {
    for(auto ptr=li.begin(); ptr!=li.end(); ptr++)scanf("%d",*ptr);
}
void in(initializer_list<ll*> li) {
    for(auto ptr=li.begin(); ptr!=li.end(); ptr++)scanf("%lld",*ptr);
}
void in(initializer_list<db*> li) {
    for(auto ptr=li.begin(); ptr!=li.end(); ptr++)scanf("%lf",*ptr);
}
void out(initializer_list<int> li) {
    for(auto ptr=li.begin(); ptr!=li.end(); ptr++)printf("%d ",*ptr);
}
void outln(initializer_list<int> li) {
    for(auto ptr=li.begin(); ptr!=li.end(); ptr++)printf("%d\n",*ptr);
}
void out(initializer_list<ll> li) {
    for(auto ptr=li.begin(); ptr!=li.end(); ptr++)printf("%lld ",*ptr);
}
void outln(initializer_list<ll> li) {
    for(auto ptr=li.begin(); ptr!=li.end(); ptr++)printf("%lld\n",*ptr);
}
int in(int &a,int &b,int &c,int &d) {
    return scanf("%d%d%d%d",&a,&b,&c,&d);
}
int in(int &a,int &b,int &c) {
    return scanf("%d%d%d",&a,&b,&c);
}
int in(int &a,int &b) {
    return scanf("%d%d",&a,&b);
}
int in(ll &a,ll &b) {
    return scanf("%lld%lld",&a,&b);
}
int in(ll &a,ll &b,ll &c,ll &d) {
    return scanf("%lld%lld%lld%lld",&a,&b,&c,&d);
}
int in(ll &a,ll &b,ll &c) {
    return scanf("%lld%lld%lld",&a,&b,&c);
}
int in(ll &a) {
    return scanf("%lld",&a);
}
int in(int &a) {
    return scanf("%d",&a);
}
int in(char *s) {
    return scanf("%s",s);
}
int in(char &c) {
    return scanf("%c",&c);
}
void out(int a,bool ln) {
    printf("%d%c",a," \n"[ln]);
}
void out(ll a,bool ln) {
    printf("%lld%c",a," \n"[ln]);
}
void out(db a,int digit,bool ln) {
    printf("%.*f%c",digit,a," \n"[ln]);
}
ll gcd(ll a, ll b) {
    return b?gcd(b,a%b):a;
}
ll lcm(ll a, ll b) {
    return a/gcd(a,b)*b;
}
const int inf = 0x3f3f3f3f;
const db eps = 1e-8;
const int maxn = 3e5+5;
const int maxm = 1e6+5;
const ll mod = 1e9+7;


int main() {
    rush() {
        int n;
        in(n);
        if(n==2){
            pf("1\n2 1\n1\n1 2\n");
            continue;
        }
        if(n==3){
            pf("5\n2 3 1\n2\n1 3\n2 3\n");
            continue;
        }
        ll ans=0;
        forrr(i,1,n-1){
            ans+=max(1LL*(i-1)*(i-1),1LL*(n-i)*(n-i));
        }
        out(ans,1);
        int tmp=(n+1)/2+1;
        out(tmp,0);
        forr(i,3,tmp){
            out(i,0);
        }
        out(1,0);
        forrr(i,tmp+1,n){
            out(i,0);
        }
        out(2,1);
        out(n-1,1);
        forr(i,2,tmp){
            out(i,0);
            out(n,1);
        }
        forrr(i,tmp,n){
            out(i,0);
            out(1,1);
        }
    }
    return 0;
}