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>
using namespace std;

long long t,n,f;
long long sq[7],curn,curs,cnt;

int main(){
    cin>>t;
    while(t--){
        cin>>n;
        curn=n;
        curs=n;
        cnt=0;
        while(!(curn==2||curn==3)){
            curs=sqrt(curn);
            curn=(curn+curs-1)/curs;
            ++cnt;
            sq[cnt]=curs;
//cout<<curs<<" "<<curn<<endl;
        }
        if(curn==2){
            cout<<n-2+cnt<<endl;
        }
        else{
            cout<<n-2+cnt+1<<endl;
        }
        for(int i=3;i<n;++i){
            f=0;
            for(int j=1;j<=cnt;++j){
                if(sq[j]==i){
                    f=1;
                }
            }
            if(f){
                continue;
            }
            else{
                cout<<i<<" "<<i+1<<endl;
            }
        }
        for(int i=1;i<=cnt;++i){
            cout<<n<<" "<<sq[i]<<endl;
            if(sq[i]!=2){
                cout<<sq[i]<<" "<<n<<endl;
            }
        }
        if(curn==2){
            cout<<"2 "<<n<<endl;
        }
        else{
            cout<<n<<" 2"<<endl;
            cout<<"2 "<<n<<endl;
        }


    }
    return 0;
}