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.io.*;
import java.util.*;

public class cf1469_Div2Edu_D {	
	public static int ceil(int x, int y) {
		return (x + y - 1) / y;
	}
	
	public static void solve(int n, PrintWriter out) {
		StringBuilder str = new StringBuilder("");
		if (n > 14) {
			int c = 0;
			for (int i = 15; i < n; i++) {
				str.append(i + " " + n + "\n");
				c++;
			}
			int cnt = 1;
			while (cnt < n) {
				str.append(n + " " + 14 + "\n");
				cnt *= 14;
				c++;
			}
			for (int i = 3; i < 14; i++) {
				str.append(i + " " + 14 + "\n");
				c++;
			}
			
			cnt = 1;
			while (cnt < 14) {
				str.append(14 + " " + 2 + "\n");
				cnt *= 2;
				c++;
			}		
			
			out.println(c);
			out.print(str);
		}
		else {
			int c = 0;
			for (int i = 3; i < n; i++) {
				str.append(i + " " + n + "\n");
				c++;
			}
			int cnt = 1;
			while (cnt < n) {
				str.append(n + " " + 2 + "\n");
				cnt *= 2;
				c++;
			}				
			
			out.println(c);
			out.print(str);
		}
	}
	
	public static void main(String args[]) throws IOException  {
		FastScanner in = new FastScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);
		int t = in.nextInt();
		for ( ; t > 0; t--) {
			int n = in.nextInt();
			solve(n, out);
		}
		out.close();
	}
	
	static class FastScanner {
	    BufferedReader br;
	    StringTokenizer st;
		
	    public FastScanner(InputStream i) {
	        br = new BufferedReader(new InputStreamReader(i));
	        st = new StringTokenizer("");
	    }
				
	    public String next() throws IOException {
	        if(st.hasMoreTokens())
	            return st.nextToken();
	        else
	            st = new StringTokenizer(br.readLine());
	        return next();
	    }

	    public int nextInt() throws IOException {
	        return Integer.parseInt(next());
	    }
	    //#
	    public long nextLong() throws IOException {
	        return Long.parseLong(next());
	    }
	    public double nextDouble() throws IOException {
	        return Double.parseDouble(next());
	    }
	    //$
	}
}
