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.*;
import java.util.*;

public class Solution {
  public static void solve(BufferedReader br) throws IOException {
    // for any number except 1,2 and 8, divide it by n, (n-4 ops)
    // then divide n by 8 until 1 (max log8(n) <= 6 ops, since (n <= 2*10**5))
    // then divide 8 by 2 until 1 (3 ops)
    // so total <= n+5 ops
    int n = Integer.parseInt(br.readLine());
    List<String> res = new ArrayList<>();
    // case 1: n < 8, divide n from 3-n-1
    // then divide n by 2 until 1
    if (n < 8) {
      for (int i=3; i<n; i++) {
        res.add(String.format("%d %d", i, n));
      }
      // divide n by 2 if n > 3 until 0
      int r = n;
      if (r > 2) {
        while (r > 1) {
          r = r/2 + (r%2==0 ? 0 : 1);
          res.add(String.format("%d %d", n, 2));
        }
      }
    } else {
      // n >= 8
      // divide 3 to n-1 (except 8) by n
      for (int i=3; i<n; i++) {
        if (i!=8) {
          res.add(String.format("%d %d", i, n));
        }
      }
      // divide n by 8 until n -> 1
      int r = n;
      if (r > 8) {
        while (r > 1) {
          r = r/8 + (r%8==0 ? 0 : 1);
          res.add(String.format("%d %d", n, 8));
        }
      }
      // divide 8 by 2 until 1
      for (int k=0; k<3; k++) {
        res.add(String.format("%d %d", 8, 2));
      }  
    }
    System.out.println(res.size());
    for (String s : res) {
      System.out.println(s);
    }
  }
  public static void main(String[] args) throws IOException {
    BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
    int t = Integer.parseInt(br.readLine());
    while (t-- > 0) {
      solve(br);
    }
  }
}
