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 hhh cerr<<"hhh"<<endl
#define see(x) cerr<<(#x)<<'='<<(x)<<endl
using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int,int> pr;
inline int read() {int x=0,f=1;char c=getchar();while(c!='-'&&c<48)c=getchar();if(c=='-')f=-1,c=getchar();while(c>47)x=x*10+(c^48),c=getchar();return f*x;}

const int maxn = 3e5+7;
const int inf = 0x3f3f3f3f;
const int mod = 1e9+7;

int a[maxn];

void solve() {
    int n=read();
    for(int i=1; i<=n; ++i) a[i]=i;
    ll ans=0;
    vector<pr> pt;
    ans+=ll(n-1)*(n-1);
    pt.push_back(pr(1,n));
    swap(a[1],a[n]);
    for(int i=2; i<=n/2; ++i) {
        ans+=2ll*(n-i)*(n-i);
        pt.push_back(pr(n-i+1,1));
        swap(a[n-i+1],a[1]);
        pt.push_back(pr(i,n));
        swap(a[i],a[n]);
    }
    if(n&1) {
        ans+=ll(n/2)*(n/2);
        pt.push_back(pr(n-n/2,1));
        swap(a[n-n/2],a[1]);
    }
    printf("%lld\n", ans);
    for(int i=1; i<=n; ++i) printf("%d%c", a[i], " \n"[i==n]);
    printf("%d\n", n-1);
    for(int i=pt.size()-1; i>=0; --i) printf("%d %d\n", pt[i].first, pt[i].second);
}

int main() {
    int T=read();
    while(T--) solve();
}