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]. 
import java.util.*;

public class ceilDivisions {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int T = sc.nextInt();
		while(T-- > 0)
		{
			int n = sc.nextInt();
			HashSet<Integer> saving = new HashSet<Integer>();
			ArrayList<Integer> save = new ArrayList<Integer>();
			int temp = n;
			saving.add(temp);
			save.add(temp);
			while(temp > 2)
			{
				float a = (float)Math.sqrt(temp);
				if(a - (int)a > 0.0)
				{
					temp = (int)a;
					temp++;
				}
				else
					temp = (int)a;
				saving.add(temp);
				save.add(temp);
			}
			int steps = n-1-saving.size() + 2*(saving.size()-1);
			System.out.println(steps);
			for(int i = 2;i < n;i++)
			{
				if(!saving.contains(i))
					System.out.println(i + " " + n);
			}
			for(int i = 0;i < save.size()-1;i++)
			{
				System.out.println(save.get(i) + " " + save.get(i+1));
				System.out.println(save.get(i) + " " + save.get(i+1));
			}
		}
	}

}
