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.awt.image.BandedSampleModel;
import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.lang.reflect.Array;
import java.util.Scanner;
public class D {

    static void sort(int[] A){

        int n = A.length;
        Random rnd = new Random();
        for(int i=0; i<n; ++i){
            int tmp = A[i];
            int randomPos = i + rnd.nextInt(n-i);
            A[i] = A[randomPos];
            A[randomPos] = tmp;
        }
        Arrays.sort(A);
    }

    public static void main(String args[])
    {
        Scanner sc=new Scanner(System.in);
        int cases = sc.nextInt();

        for(int i=0;i<cases;i++)
        {
            int n = sc.nextInt();
            long two[]=new long[]{1,2,4,16,256,65336};




            int c=0;
            StringBuilder ans=new StringBuilder();
            for(int j=1;j<=n-1;j++)
            {
                if(Arrays.binarySearch(two,j)<0)
                {
                    ans.append((j)+" "+(j+1)+"\n");
                    c++;
                }
            }

            if(n>65336)
            {
                c=c+2;
                ans.append(n+" "+65336+"\n");
                ans.append(n+" "+65336+"\n");
                n=65336;
            }

            if(n>256)
            {
                c=c+2;
                ans.append(n+" "+256+"\n");
                ans.append(n+" "+256+"\n");
                n=256;
            }

            if(n>16)
            {
                c=c+2;
                ans.append(n+" "+16+"\n");
                ans.append(n+" "+16+"\n");
                n=16;
            }

            if(n>4)
            {
                c=c+2;
                ans.append(n+" "+4+"\n");
                ans.append(n+" "+4+"\n");
                n=4;
            }

            if(n>2)
            {
                c=c+2;
                ans.append(n+" "+2+"\n");
                ans.append(n+" "+2+"\n");
                n=2;
            }

            System.out.println(c);
            System.out.print(ans);

        }

    }
}