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<vector>
using namespace std;
int ceil(int x,int y){
    return (x+y-1)/y;
}
int main(){
    int t;
    cin>>t;
    while(t--){
        int n;
        cin>>n;
        
        int count=0;
        //start from n cancel all numbers until it reaches n/2
        //root(n) wil cancel n
        vector<int> v;
        int x=2,y=3;
        v.push_back(x);
        while(y<=n){
            if(ceil(y,x)==x){
                v.push_back(y);
                count+=(y-x+1);
                x=y;
            }
            y++;
        }
        count+=(n-x+1);
        if(n==x){
            count--;
        }
        cout<<count<<endl;
        int l;
        while(!v.empty()){
            l=v.back();v.pop_back();
            if(n==l){
                l=v.back();v.pop_back();
            }
            //x-y++1 steps
            for(int i=n-1;i>=(l+1);i--){
                cout<<i<<" "<<n<<endl;
            }
            // if(n==l){
            //     cout<<n<<" "<<l<<endl;
            // }
            // else{
            cout<<n<<" "<<l<<endl;
            cout<<n<<" "<<l<<endl;
            // }

            n=l;
        }
    }
    
}