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 fo(a,b) for(int a=0;a<b;++a)
#define pb push_back
#define printList(v) for(auto _x:(v)) cout<<_x<<" ";cout<<endl;

////////////// DEBUG UTILS //////////////////
#define debug(...) __print(#__VA_ARGS__, __VA_ARGS__)
template<class K,class V> ostream& operator<<(ostream& os,const pair<K,V>& p) {os<<"("<<p.first<<", "<<p.second<<")"; return os;}
template<class T> ostream& __dPrint(ostream& os, const T& v) {os<<v; return os;}
template<> ostream& __dPrint(ostream& os, const string& v) {os<<'"'<<v<<'"'; return os;}
template<class T> ostream& operator <<(ostream& os,const vector<T>& v) {
  os<<v.size()<<" vals: [ ";
	fo(i,v.size()) { if(i)os<<", "; __dPrint(os, v[i]); } os<<" ]";
	return os;
}
template <typename Arg> void __print(const char* name, Arg&& arg){ cerr << name << " : " << arg << endl; }
template <typename Arg, typename... Args>
void __print(const char* names, Arg&& arg, Args&&... args){
	const char* comma = strchr(names + 1, ',');
  cerr.write(names, comma - names) << " : " << arg<< endl;
  __print(comma+1, args...);
}
////////////// END OF DEBUG UTILS //////////////////
 
typedef long long int lint;
typedef vector<string> VS;
typedef vector<int> VI;
typedef vector<lint> VL;
typedef vector<VI> VVI;
typedef vector<VL> VVL;
typedef pair<int,int> pii;
 
int ri() {int x;if(scanf(" %d",&x) == EOF) return -1;return x;}
lint rl() {lint x;if(cin>> x) return x; return -1;}
string rs() {string s;if(cin>>s) return s; return "";}
/////////////////////////// END TEMPLATE /////////////////
 
struct SingleCase {
  int n;

  bool solveCase() {
    n = ri(); if(n < 0) return false;
    vector<pii> kills;
    
    lint score = 0;
    for(int i=1;i<n;i++) {
      int a = i, b = n-1-i;
      int t = max(a,b);      
      if(a > b) {
        kills.pb({0, i});
      } else {
        kills.pb({n-1, i});
      }
      score += (lint)t*t;
    }

    VI ret(n, -1);
    VI head(n); fo(i,n) head[i] = i;
    VI destroyed(n, 0);
    for(const auto& [a, b] : kills) {
      ret[head[a]] = b;
      head[a] = head[b];
      destroyed[b] = 1;
    }
    fo(i,n) if(!destroyed[i]) {
      ret[head[i]] = i;
      break;
    }

    cout << score << endl;

    fo(i,n) cout << ret[i] + 1 <<" "; cout << endl;
    cout << kills.size() << endl;
    for(const auto& [a, b]: kills) {
      cout << b+1 << " " << a + 1 << endl;
    }

    return true;
  }
};
 
int main() {
  ri();
  while(SingleCase().solveCase()) {};
  return 0;
}