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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintStream;
import java.util.*;

public class D {
    static HashMap<Long, Long> combos = new HashMap<>();
    public static void main(String[] args) throws IOException {
        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        PrintStream out = System.out;

        int b = Integer.parseInt(br.readLine());

        for (int i = 0; i < b; i++) {
            int n = Integer.parseInt(br.readLine());
            ArrayList<Integer[]> ans = new ArrayList<>();
            boolean[] keep = new boolean[n+1];
            ArrayList<Integer> kept = new ArrayList<>();
            while (n > 2) {
                keep[n] = true;
                kept.add(n);
                n = (int)(Math.sqrt(n)+0.999999);
            }
            int m = 0;
            n = keep.length-1;
            for (int x = 3; x < n; x++) {
                if (keep[x]) continue;
                ans.add(new Integer[]{x, x+1});
                m++;
            }
            for (int x = 0; x < kept.size(); x++) {
                int curr = kept.get(x);
                int next = 2;
                if (x+1 < kept.size()) next = kept.get(x+1);
                ans.add(new Integer[]{curr, next});
                ans.add(new Integer[]{curr, next});
                m += 2;
            }
            out.println(m);
            for (Integer[] x : ans) {
                out.println(x[0] + " " + x[1]);
            }
        }

        System.out.flush();
    }
}