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.io.*;
import java.util.*;
public class Ishu
{
    static Scanner scan = new Scanner(System.in);
    static int m = 0;
    static int[][] ans = new int[1][2];
    static void solve(int n)
    {
    if(n <= 2)
        return;
        
    int half = (int)Math.ceil(Math.sqrt(n));
/*    if(n % 2 == 1)
        ++half;*/
        
    int i;
    for(i=half+1;i<n;++i)
        {
        ans[m][0] = i;
        ans[m++][1] = n;
        }
    ans[m][0] = n;
    ans[m++][1] = half;
    
    ans[m][0] = n;
    ans[m++][1] = half;
    
    solve(half);
    }
   
    static void tc()
    {
    int n = scan.nextInt();
    int i;
    
    m = 0;
    ans = new int[2*n][2];
    solve(n);
    System.out.println(m);
    for(i=0;i<m;++i)
        System.out.println(ans[i][0] + " " + ans[i][1]);
    System.out.println();
    }
    public static void main(String[] args)
    {
    int t = 1;
    t = scan.nextInt();
    while(t-- > 0)
        tc();
    }
}

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