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]. 
#include <iostream>
#include<bits/stdc++.h>
#define li long long int
using namespace std;

int main()
{
    int t;
    cin>>t;

    while(t--){

        li n;
        cin>>n;

        vector<pair<li,li> > ans;

        // appproach --> optimal soln --> we keep 1 and 2 as they are
        // change the rest n-2 numbers with 1
        // numbers from 3 to n-1 can be easily changed to 1 in (n-4) operations
        // now the question would to reduce n to 1
        // therefore try and reduce n at every step when converting numbers from (n-1 to 3) to 1

        li temp=n;

        for(int i=(n-1);i>=3;i--){

            // check if reducing n to n/i is optimal

            // ceil value --> (n+i-1)/i

            if( ((temp+i-1)/i)>=i ){

                ans.push_back(make_pair(n,i));
                temp=(temp+i-1)/i;

            }

            // compulsory step --> of converting i to 1

            ans.push_back(make_pair(i,n));

        }

        if(temp>1){

            while(temp>1){
                ans.push_back(make_pair(n,2));
                temp=(temp+1)/2;

            }


        }

        if(ans.size()<=(n+5)){

            int nn=ans.size();

            cout<<nn<<endl;

            for(int i=0;i<nn;i++){

                cout<<ans[i].first<<" "<<ans[i].second<<endl;;

            }

        }


    }

    return 0;
}
