#include<bits/stdc++.h>
using namespace std;
#define int long long
#define fast_io ios_base::sync_with_stdio(0);cin.tie(0) 
#define mod 1e9 + 7
#define pb push_back
#define vi vector<int>


int32_t main(){
    fast_io;
    int t;cin>>t;
    while(t--){
     int n;cin>>n;
     vector<pair<int,int>> ans;
     int a = n;
     vi special;
     set<int> s;
     special.pb(a);
     s.insert(a);
     a=sqrt(a);
     
     while(a>=2){
      special.pb(a+1);
      s.insert(a+1);
      a = sqrt(a);
     }
     /*for(int i=0;i<special.size();i++){
      cout<<special[i]<<" ";
     }*/
     for(int i=3;i<=n;i++){
      if(s.find(i) != s.end()){
          continue;
      }
      ans.pb({i,i+1});
     }
     
     for(int i=0;i<special.size()-1;i++){
      //cout<<"satyam"<<" ";
      ans.pb({special[i],special[i+1]});
      ans.pb({special[i],special[i+1]});
     }
     ans.pb({special[special.size()-1],2});
     ans.pb({special[special.size()-1],2});
     cout<<ans.size()<<endl;
     for(int i=0;i<ans.size();i++){
         cout<<ans[i].first<<" "<<ans[i].second<<endl;
     }
    }
    return 0;
}
### Prompt

Develop a solution in CPP to the problem described below:

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]. 

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
#define int long long
#define fast_io ios_base::sync_with_stdio(0);cin.tie(0) 
#define mod 1e9 + 7
#define pb push_back
#define vi vector<int>


int32_t main(){
    fast_io;
    int t;cin>>t;
    while(t--){
     int n;cin>>n;
     vector<pair<int,int>> ans;
     int a = n;
     vi special;
     set<int> s;
     special.pb(a);
     s.insert(a);
     a=sqrt(a);
     
     while(a>=2){
      special.pb(a+1);
      s.insert(a+1);
      a = sqrt(a);
     }
     /*for(int i=0;i<special.size();i++){
      cout<<special[i]<<" ";
     }*/
     for(int i=3;i<=n;i++){
      if(s.find(i) != s.end()){
          continue;
      }
      ans.pb({i,i+1});
     }
     
     for(int i=0;i<special.size()-1;i++){
      //cout<<"satyam"<<" ";
      ans.pb({special[i],special[i+1]});
      ans.pb({special[i],special[i+1]});
     }
     ans.pb({special[special.size()-1],2});
     ans.pb({special[special.size()-1],2});
     cout<<ans.size()<<endl;
     for(int i=0;i<ans.size();i++){
         cout<<ans[i].first<<" "<<ans[i].second<<endl;
     }
    }
    return 0;
}
```