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]. 
//package eround101;

import java.util.Scanner;

public class C101 {

    static Scanner sc = new Scanner(System.in);

    static void solve() {

        int n = sc.nextInt();
        int m = n, o = n;
        int count = 0;
        if (n > 8) {
            while (n != 1) {
                if (n % 8 == 0) {
                    n = n / 8;
                } else {
                    n = n / 8 + 1;
                }
                count++;
            }
            System.out.println(o - 1 + count);
            for (int i = m; i >= 4; i--) {
                if (i == 9) {
                    continue;
                }
                System.out.println(i - 1 + " " + o);
            }
            while (m != 1) {
                System.out.println(o + " " + 8);
                if (m % 8 == 0) {
                    m = m / 8;
                } else {
                    m = m / 8 + 1;
                }
                
            }
             System.out.println(8 + " " + 2);
             System.out.println(8 + " " + 2);
             System.out.println(8 + " " + 2);
             
            return;
        } else {
            if (n == 3) {
                System.out.println(n - 1);
                System.out.println(3 + " " + 2);
                System.out.println(3 + " " + 2);
                return;
            } else if (n == 4) {
                System.out.println(n - 1);
                for (int i = m; i >= 4; i--) {
                    System.out.println(i - 1 + " " + i);

                }
                while (m != 1) {
                    System.out.println(o + " " + 2);
                    if (m % 2 == 0) {
                        m = m / 2;
                    } else {
                        m = m / 2 + 1;
                    }

                }
                return;

            } else {
                System.out.println(n);
                for (int i = m; i >= 4; i--) {
                    System.out.println(i - 1 + " " + o);

                }
                while (m != 1) {
                    System.out.println(o + " " + 2);
                    if (m % 2 == 0) {
                        m = m / 2;
                    } else {
                        m = m / 2 + 1;
                    }

                }
                return;
            }
        }
    }

    public static void main(String[] args) {

        int t = sc.nextInt();

        while (t-- > 0) {
            solve();
        }

    }
}
