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 JAVA):

Solution:
import java.util.*;
import java.io.*;

public class Question4 {

	static Scanner sc =new Scanner(System.in);
	static BufferedWriter bw = new BufferedWriter(new OutputStreamWriter(System.out));
	
	public static void main(String[] args) throws IOException {
		int t = sc.nextInt();
		while(t-->0) {
			solve();
		}
		bw.flush();
		bw.close();
	}
	
	public static void solve() throws IOException {
		
		int n= sc.nextInt();
		
		int val = 2;
		
		if(n >= 8)val = 8;
		else if(n >= 4)val = 4;
		
		if(n == 3) {
			print();
			return;
		}
		
		HashSet<Integer> set = new HashSet<Integer>();
		set.add(1);
		set.add(2);
		set.add(n);
		set.add(val);
		
		//System.out.println(set);
		
		int count = n - set.size();
		int atemp = n;
		while(atemp != 1) {
			count++;
			if(atemp % val == 0)atemp = atemp/val;
			else atemp = atemp/val + 1;
		}
		
		//second 
		int btemp = val;
		while(btemp > 1) {
			count++;
			btemp = btemp/2;
		}
		if(n == 4 || n == 8)count--;
		bw.write(String.valueOf(count) + "\n");
		
		if(count > n + 5) {
			System.out.println("WRONG!!!");
			return;
		}
		
		for(int i = 1;i < n;i++) {
			if(set.contains(i)) {
				continue;
			}
			bw.write(String.valueOf(i) + " " + String.valueOf(n) + "\n");
		}
		
		int temp = n;
		if(temp != val) {
			
			while(temp > 1) {
				bw.write(String.valueOf(n) + " " + String.valueOf(val) + "\n");
				
				if(temp % val == 0)temp = temp/val;
				else temp = temp/val + 1;
				
			}
			
		}
		

		int valv = val;
		while(val > 1) {
			
			bw.write(String.valueOf(valv) + " " + String.valueOf(2) + "\n");
			val = val/2;
		
		}
		
	}
	
	public static void print() throws IOException {
	
		bw.write("2\n3 2\n3 2\n");
		
	}
}