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>
#define uint unsigned int
#define int long long int
#define ld long double
#define INF 1e9
using namespace std;
const int mod=1e9+7;

int32_t main()
{  
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
  
  int t;
  cin>>t;
//  t=1;
  while(t--)
  {
      int n;
      cin>>n;
     int a[7]={1,2,3,5,8,13,21};
     int x=n,c=0;
     while(x>1)
     {
         x/=2;
         c++;
     }
     if(n==3)
     {
         cout<<"2\n3 2\n3 2\n";
         continue;
     }
     if(n==4)
     {
         cout<<"3\n3 4\n4 2\n4 2\n";
         continue;
     }
     int i,j,k,p=0;
     for(i=6;i>=0;i--)
     {
         if(c>=a[i])
         {
             p=i;
             break;
         }
     }
// cout<<c<<" "<<p<<" ";
     
     if(n==(1LL<<a[p]))
     {
         cout<<(n+4)<<"\n";
     }
     else
     cout<<(n+5)<<"\n";
     for(i=2;i<n;i++)
     {
         if(i==(1LL<<a[p]) || i==(1LL<<a[p-1]))
         continue;
         else
         cout<<i<<" "<<n<<"\n";
     }
     if(n!=(1LL<<a[p]))
     {
     cout<<n<<" "<<(1LL<<a[p])<<"\n";
     cout<<n<<" "<<(1LL<<a[p-1])<<"\n";
     }
     for(i=0;i<=6;i++)
     {
         if(i%2==0)
         {
             cout<<(1LL<<a[p])<<" "<<(1LL<<a[p-1])<<"\n";
         }
         else
         cout<<(1LL<<a[p-1])<<" "<<(1LL<<a[p])<<"\n";
     }
  }
}