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.BufferedOutputStream;
import java.io.PrintStream;
import java.util.Scanner;


public class Solution {

    private static boolean TESTS = true;

    private final Scanner in;
    private final PrintStream out;

    public Solution(final Scanner in, final PrintStream out) {
        this.in = in;
        this.out = out;
    }


    public void solve() {
        for (int tests = TESTS ? in.nextInt() : 1; tests > 0; tests--) {
            final int n = in.nextInt();
            final int[] split = split(n);

            if (split == null) {
                out.println(n + 5);

                for (int i = 3; i < n; i++) {
                    out.println(i + " " + (i + 1));
                }

                for (int i = 0; i < 8; i++) {
                    out.println(n + " " + 2);
                }
            } else {
                out.println(n + 5);

                for (int i = 3; i < n; i++) {
                    if(i != split[0] && i != split[1])
                    out.println(i + " " + (i + 1));
                }

                out.println(n + " " + split[0]);
                out.println(split[0] + " " + n);

                out.println(n + " " + split[1]);
                out.println(split[1] + " " + n);

                for (int i = 0; i < 6; i++) {
                    out.println(n + " " + 2);
                }
            }

        }
    }

    private static int[] split(final int n) {
        if (n > 128) {
            final int x = (int) Math.sqrt(n - 1);
            final int y = (n +x - 1) / x;
            final int z = (int) Math.sqrt(y - 1);

            return new int[]{x, z};
        }

        return null;
    }

    public static void main(final String[] args) {
        final Scanner in = new Scanner(System.in);
        final PrintStream out = new PrintStream(new BufferedOutputStream(System.out));

        try {
            new Solution(in, out).solve();
        } finally {
            out.flush();
        }
    }

}
