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<iostream>
#include<math.h>
using namespace std;
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
         int n,x=1;
    cin>>n;
    if(n==3)
    {
        cout<<2<<endl<<3<<" "<<2<<endl<<3<<" "<<2<<endl;
        continue;
    }
if(n>10)
{
    while(pow(8,x)<n)
   x++;
  cout<<n-1+x<<endl;
    for(int i=3;i<=7;i++)
    cout<<i<<" "<<i+1<<endl;
    for(int i=9;i<=n-1;i++)
    cout<<i<<" "<<i+1<<endl;
    while(x--)
   cout<<n<<" "<<8<<endl;
    cout<<8<<" "<<2<<endl<<8<<" "<<2<<endl<<8<<" "<<2<<endl;  
    continue;
}
   while(pow(3,x)<n)
   x++;
   cout<<n-4+2+x<<endl;
   for(int i=4;i<=n-1;i++)
   cout<<i<<" "<<i+1<<endl;
   while(x--)
   cout<<n<<" "<<3<<endl;
    cout<<3<<" "<<2<<endl<<3<<" "<<2<<endl;   
    }

     return 0;
}