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;

int main(){
    ios::sync_with_stdio(false);
    cin.tie(0);
    int t;
    cin >> t;
    for(int q =0; q< t; ++q){
        
        double n;
        cin >> n;
        double a = ceil(ceil(log2(n))/2.0);
        //cout << a;
        deque<pair<int,int>>p;
        set<double>s;
        for(double i = 3;i<n+1;++i)s.insert(i);
        p.push_back(make_pair(n,pow(2.0,a)));
        p.push_back(make_pair(n,pow(2.0,a)));
        s.erase(n);
        //a = ceil(a/2.0);
        while(a!=1  || a>1){
            double x = pow(2.0,a);
            double y = pow(2.0,ceil(a/2.0));
            p.push_back(make_pair(x,y));
            p.push_back(make_pair(x,y));
            s.erase(x);
            a = ceil(a/2.0);
        }
        double pivot = n;
        for(auto it = s.begin();it!=s.end();++it){
            //cout << "---"<<*it<<"----";
            p.push_front(make_pair(*it,n));
        }
        int len = p.size();
        cout << len <<"\n";
        while(!p.empty()){
            cout << p.front().first <<" "<<p.front().second<<"\n";
            p.pop_front();
        }


    }
    return 0;
}