Problem: You have an array a_1, a_2, ..., a_n where a_i = i.

In one step, you can choose two indices x and y (x ≠ y) and set a_x = \left⌈ (a_x)/(a_y) \right⌉ (ceiling function).

Your goal is to make array a consist of n - 1 ones and 1 two in no more than n + 5 steps. Note that you don't have to minimize the number of steps.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first and only line of each test case contains the single integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the length of array a.

It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print the sequence of operations that will make a as n - 1 ones and 1 two in the following format: firstly, print one integer m (m ≤ n + 5) — the number of operations; next print m pairs of integers x and y (1 ≤ x, y ≤ n; x ≠ y) (x may be greater or less than y) — the indices of the corresponding operation.

It can be proven that for the given constraints it's always possible to find a correct sequence of operations.

Example

Input


2
3
4


Output


2
3 2
3 2
3
3 4
4 2
4 2

Note

In the first test case, you have array a = [1, 2, 3]. For example, you can do the following: 

  1. choose 3, 2: a_3 = \left⌈ (a_3)/(a_2) \right⌉ = 2 and array a = [1, 2, 2]; 
  2. choose 3, 2: a_3 = \left⌈ 2/2 \right⌉ = 1 and array a = [1, 2, 1]. 

You've got array with 2 ones and 1 two in 2 steps.

In the second test case, a = [1, 2, 3, 4]. For example, you can do the following: 

  1. choose 3, 4: a_3 = \left⌈ 3/4 \right⌉ = 1 and array a = [1, 2, 1, 4]; 
  2. choose 4, 2: a_4 = \left⌈ 4/2 \right⌉ = 2 and array a = [1, 2, 1, 2]; 
  3. choose 4, 2: a_4 = \left⌈ 2/2 \right⌉ = 1 and array a = [1, 2, 1, 1]. 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#include <unordered_map>
#include <cctype>
#include <string.h>

#define PB push_back
#define EB emplace_back 
#define MP make_pair
#define endl "\n"
#define ll long long
#define ull unsigned long long
#define STR string
#define VEC vector<int>
#define IOS {ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);}
#define loop(i,a,b,c) for(ll i=a;i<b;i+=c)
#define MOD (int)1000000007
#define sp " "
#define pii pair<int,int>


using namespace std;
using namespace std::chrono;

/* Debugging */
template <typename T>
void printarr(T *A, int n){
  for (int i = 0; i < n; i++)
  {
    cout<<A[i]<<" ";
  }
  cout<<endl;  
}
template <typename T>
void arrinit(T *A, int n, T data){
  for (int i = 0; i < n; i++)
  {
    A[i] = data;
  }
  
}
template <typename T>
void takeinput(T *A, int n){
  for(int i=0;i<n;i++){
    cin>>A[i];
  }
}
template <typename T>
void printcont(T A){
  for(auto a: A){
    cout<<a+1<<" ";
  }
  cout<<endl;
}
template <typename T>
T binpowmod(T a, T b, T p){
  if(b==0) return 1;
  if(b==1) return a;
  if(b%2==0) return ((binpowmod(a,b/2,p)%p)*(binpowmod(a,b/2,p)%p))%p;
  return ((a%p)*(binpowmod(a,b-1,p)%p))%p;
}
// template <typename T>
// T gcd(T a,T b){
//   if(a==0) return b;
//   if(b==0) return a;
//   if(a>b) return gcd(b,a%b);
//   return gcd(b%a,a); 1 2 3 4  => 1 2 
// }

void solve (){
  int n;
  cin>>n;
  int k = n;
  vector <pii> steps{};
  if(n<=8){
    for (int i = 3; i < n; i++)
    {
        steps.PB({i,i+1});
    }
    while(n!=1){
        steps.PB({k,2});
        if(n%2==1) n = n/2 + 1;
        else n/=2;
    }  
  }
  else{
    for (int i = 3; i < n; i++)
    {
        if(i!=8)steps.PB({i,i+1});
    }
    while(n!=1){
        steps.PB({k,8});
        if(n%8!=0) n = n/8 + 1;
        else n = n/8;
    }
    steps.PB({8,2});
    steps.PB({8,2});
    steps.PB({8,2});
  }
  cout<<steps.size()<<endl;
  for (int i = 0; i < steps.size(); i++)
  {
      cout<<steps[i].first<<" "<<steps[i].second<<endl;
  }
  
  
}

int main(){
    auto start = high_resolution_clock::now(); 
    IOS
    // freopen("input.txt","r",stdin);
    // freopen("output.txt","w",stdout);
    // freopen("output.txt","w",stderr);
    ll t;
    cin>>t;  
    while(t--){
      solve();
    }
    auto stop = high_resolution_clock::now(); 
    auto duration = duration_cast<milliseconds>(stop - start); 
    // cout <<"Time taken : "<<duration.count() <<" milliseconds\n";
    return 0;
}