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;
void solve(){
	int n;
	cin>>n;
	int fake=n-1;
	int count=0;
	vector<pair<int,int> > ve;
	while(fake>32){
		ve.push_back(make_pair(n,fake));
		fake--;
	}
	if(fake==32){
		double temp=n;
		while(temp>1){
			ve.push_back(make_pair(fake,n));
			temp=ceil(temp/fake);
		}
	}
	else
	fake=n;
	int just=fake-1;
	while(just>2){
		ve.push_back(make_pair(fake,just));
		just--;
	}
	just=fake;
	double ans=fake;
	while(ans>1){
		ve.push_back(make_pair(2,just));
		double le=ans/2;
		//cout<<le<<endl;
		ans=ceil(le);
	}
	int len=ve.size();
	cout<<len<<endl;
	for(int i=0;i<len;i++){
		cout<<ve[i].second<<" "<<ve[i].first<<endl;
	}
}
int main(void){
	int t;
	cin>>t;
	while(t--){
		solve();
	}
}