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>
#define ll long long
#define pb push_back
#define pll pair<long long, long long>
#define f first
#define s second
#define debug cout<<"here\n" 
#define MOD 1000000007

const int MAXN = 2e5 +15;
using namespace std;



void solve(){	
	int n,m; cin>>n;
	m = n;
	vector<pair<int,int> > v;
	while(m>3){
		n=m;
		for(ll i=n-1; i*i>=n; i--){
			m = i;
			if((i-1)*(i-1) < n) break;
			v.pb({i,n});
		}
		v.pb({n,m}); v.pb({n,m});
	}
	if(m == 3){
		v.pb({3,2});
		v.pb({3,2});
	} 
	cout<<v.size();
	for(auto pp: v){
		cout<<"\n"<<pp.f<<" "<<pp.s;
	}


}
 
int main(){
	ios_base::sync_with_stdio(false);
	cout.tie(0); cin.tie(0);	
	cout<<fixed<<setprecision(12);

	int t =1; cin>>t;
	for(int i=1; i<= t; i++){
		solve();
		cout<<"\n";
	}
	
	return 0;
}