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.*;
import java.io.*;
 
public class Main{
    public static void main(String[] args) throws IOException{
    	In in = new In();
    	for(int qr = in.pint(); qr > 0; qr--) {
    		int n = in.pint();
    		StringBuilder sb = new StringBuilder();
    		int m = 0;
    		for(int i = 3; i < n; i++) {
    			if(i == 8) {continue;}
    			m++;
    			sb.append(i);
    			sb.append(' ');
    			sb.append(i + 1);
    			sb.append('\n');
    		}
    		int k = n;
    		if(k > 8) {
    			while(k > 1) {
    				m++;
        			sb.append(n);
        			sb.append(" 8\n");
        			k = (int)(Math.ceil(k / 8.0));
        		}
    			for(int i = 0; i < 3; i++) {
    				m++;
    				sb.append("8 2\n");
    			}
    		}else {
    			while(k > 1) {
    				m++;
    				sb.append(n);
    				sb.append(" 2\n");
        			k = (int)(Math.ceil(k / 2.0));
    			}
    		}
    		System.out.println(m);
    		System.out.print(sb);
    	}
    }
}   
class In{
    BufferedReader in;
    StringTokenizer st = new StringTokenizer("");
    public In(){
        in = new BufferedReader(new InputStreamReader(System.in));
    }
    
    int pint() throws IOException{
        if(st.hasMoreTokens()) {return Integer.parseInt(st.nextToken());}
        else {return Integer.parseInt(in.readLine());}
    } 
    double pdbl() throws IOException{
        if(st.hasMoreTokens()) {return Double.parseDouble(st.nextToken());}
        else {return Double.parseDouble(in.readLine());}
    }
    long plng() throws IOException{
        if(st.hasMoreTokens()) {return Long.parseLong(st.nextToken());}
        else {return Long.parseLong(in.readLine());}
    }
    char pchr() throws IOException{
        if(st.hasMoreTokens()) {return st.nextToken().charAt(0);}
        else {return in.readLine().charAt(0);}
    }
    String pstr() throws IOException{
        if(st.hasMoreTokens()) {return st.nextToken();}
        else {return in.readLine();}
    }
    String readLine() throws IOException{
        return in.readLine();
    }
    boolean ready() throws IOException {return in.ready();}
    boolean readyN() throws IOException{return in.ready() || st.hasMoreTokens();}
    boolean tokens() {return st.hasMoreTokens();}
    void tok() throws IOException{st = new StringTokenizer(in.readLine());}
    void skip() throws IOException{in.readLine();}
}