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 D{
	
	public static void main(String[] args){
		Scanner in = new Scanner(System.in);
		StringBuilder res = new StringBuilder();
		int t = in.nextInt();
		o: for(int g = 0;g<t;++g){
			long n = in.nextInt();
			StringBuilder localRes = new StringBuilder();
			long count = 0;
			if(n < 33){
				for(int i = 3;i<=n-1;++i){
					localRes.append(i + " " + n + "\n");
				}
				count = n-3;
				long localN = n;
				while(localN > 1){
					count++;
					localN = localN / 2 + (localN / 2 * 2< localN ? 1 : 0);
					localRes.append(n + " " + 2 + "\n");
				}
			} else{
				for(int i = 3;i<=n-1;++i){
					if(i == 32)
						continue;
					localRes.append(i + " " + n + "\n");
				}
				count = n-4;
				long localN = n;
				while(localN > 1){
					count++;
					localN = localN / 32 + (localN / 32 * 32< localN ? 1 : 0);
					localRes.append(n + " " + 32 + "\n");
				}
				count += 5;
				for(int i = 1;i<=5;++i){
					localRes.append(32 + " " + 2 + "\n");
				}
			}
			res.append(count + "\n" + localRes.toString());
		}
		System.out.println(res.toString());
		
		
	}
}