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:
//mangesh2102000

#include <bits/stdc++.h>
using namespace std;

#define IOS ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL)
#define int long long int 
#define ld long double
#define pb push_back
#define tr(c,it) for(auto it = (c).begin(); it != (c).end(); ++it)
#define F first
#define S second
#define mod 1000000007  


void solve(int t)
{
	int n;
	cin >> n;

	if(n >= 9)
	{
		cout << n+5 << "\n";
		int i=3,j=4,itr=n-3;
		while(itr--)
		{
			if(i != 8)
				cout << i << " " << j << "\n";
			i++;
			j++;
		}
		cout << n << " " << 8 << "\n";
		cout << n << " " << 8 << "\n";
		cout << n << " " << 8 << "\n";
		cout << n << " " << 8 << "\n";
		cout << n << " " << 8 << "\n";
		cout << n << " " << 8 << "\n";
		cout << 8 << " " << 2 << "\n";
		cout << 8 << " " << 2 << "\n";
		cout << 8 << " " << 2 << "\n";

	}
	else
	{
		if(n == 3)
		{
			cout << n-1 << "\n";
			cout << n << " " << 2 << "\n"; 
			cout << n << " " << 2 << "\n"; 
		}
		else if(n == 4)
		{
			cout << n-1 << "\n";
			cout << 3 << " " << 4 << "\n"; 
			cout << n << " " << 2 << "\n"; 
			cout << n << " " << 2 << "\n"; 
		}
		else if(n == 5)
		{
			cout << n << "\n";
			cout << 3 << " " << 4 << "\n"; 
			cout << 4 << " " << 5 << "\n";
			cout << n << " " << 2 << "\n"; 
			cout << n << " " << 2 << "\n"; 
			cout << n << " " << 2 << "\n"; 

		}
		else if(n == 6)
		{
			cout << n << "\n";
			cout << 3 << " " << 4 << "\n"; 
			cout << 4 << " " << 5 << "\n"; 
			cout << 5 << " " << 6 << "\n";
			cout << n << " " << 2 << "\n"; 
			cout << n << " " << 2 << "\n"; 
			cout << n << " " << 2 << "\n"; 
		}
		else if(n == 7)
		{
			cout << n << "\n";
			cout << 3 << " " << 4 << "\n"; 
			cout << 4 << " " << 5 << "\n"; 
			cout << 5 << " " << 6 << "\n"; 
			cout << 6 << " " << 7 << "\n"; 
			cout << n << " " << 2 << "\n"; 
			cout << n << " " << 2 << "\n"; 
			cout << n << " " << 2 << "\n"; 
		}
		else if(n == 8)
		{
			cout << n << "\n";
			cout << 3 << " " << 4 << "\n"; 
			cout << 4 << " " << 5 << "\n"; 
			cout << 5 << " " << 6 << "\n"; 
			cout << 6 << " " << 7 << "\n"; 
			cout << 7 << " " << 8 << "\n"; 
			cout << n << " " << 2 << "\n"; 
			cout << n << " " << 2 << "\n"; 
			cout << n << " " << 2 << "\n"; 
		}
	}
    //cout << "Case #" << t << ": "; 
	
}   

    
int32_t main()
{
    IOS;
    int t;
    cin >> t;
    for(int i=1;i<=t;i++)
    {
        solve(i);
    }

    //cerr<< '\n' << "Time elapsed :" << clock() * 1000.0 / CLOCKS_PER_SEC << " ms\n" ;

    return 0;
}