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:


//@author->.....future_me......//

//..............Learning.........//

import java.util.*;
import java.io.*;
import java.lang.*;
import java.math.BigInteger;

public class B {

  // >>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Fast IO <<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<< //

  static class FastReader {
    BufferedReader  br;
    StringTokenizer st;

    public FastReader() {
      br = new BufferedReader(new InputStreamReader(System.in));
    }

    String next() {
      while (st == null || !st.hasMoreElements()) {
        try {
          st = new StringTokenizer(br.readLine());
        } catch (IOException e) {
          e.printStackTrace();
        }
      }
      return st.nextToken();
    }

    int nextInt() {
      return Integer.parseInt(next());
    }

    long nextLong() {
      return Long.parseLong(next());
    }

    double nextDouble() {
      return Double.parseDouble(next());
    }

    String nextLine() {
      String str = "";
      try {
        str = br.readLine();
      } catch (IOException e) {
        e.printStackTrace();
      }
      return str;
    }
  }

  // >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Code Starts <<<<<<<<<<<<<<<<<<<<<<<<<<<<< //

  static FastReader  sc  = new FastReader();
  static PrintWriter out = new PrintWriter(System.out);

  static int mod = 1000000007;

  public static void main(String[] args) throws Exception {
//    try {
//
    int t = sc.nextInt();
    while (t-- > 0)
      B.go();
    out.flush();
//    } catch (Exception e) {
//      return;
//    }
  }

  static void go() {
    int n = sc.nextInt();
    int nn = n;
    int x = 0;
    if (n > 8) {
      x = 8;
//      if (n % x == 0) {
//        x++;
//      }
    } else
      x = 2;

    if (n == 3) {
      System.out.println(2);
      System.out.println(3 + " " + 2);
      System.out.println(3 + " " + 2);
      return;
    }
    ArrayList<dis> aa = new ArrayList<>();

    int j = n - 1;
    while (j > 2) {
      if (j != x)
        aa.add(new dis(j, n));

      j--;

    }

    int t1 = n;
    int t2 = x;

    while (t1 > 1) {
      aa.add(new dis(n, x));
      t1 = (t2 + t1 - 1) / t2;

    }
//    System.out.println(t2 + " " + t1 + " " + x + " " + n);
    if (t2 > 2) {
      while (t2 >= 2) {
        aa.add(new dis(x, 2));
        t2 = (t2 + 1) / 2;
      }
    }

    if (nn <= 10) {
      if (x != 2) {
        aa.add(new dis(x, 2));
        aa.add(new dis(x, 2));
      }
    }
    System.out.println(aa.size());
    for (dis i : aa) {
      System.out.println(i.x + " " + i.y);
    }
    System.out.println();
//    System.out.println(t2 + " " + t1 + " " + x + " " + n + " " + aa.size());

  }

  static long dp[] = new long[15];

  static long solve(int x[], int l, int r, int n) {
    long xx = 0;
    if (l < n)

      xx = x[l];
    else {
      return Integer.MAX_VALUE;
    }
    for (int i = l + 1; i <= r && i < n; i++) {
      xx ^= x[i];
    }
    return xx;
  }

  // >>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> Code Ends <<<<<<<<<<<<<<<<<<<<<<<<<<<<< //

  static void sort(long[] a) {
    ArrayList<Long> aa = new ArrayList<>();
    for (long i : a) {
      aa.add(i);
    }
    Collections.sort(aa);
    for (int i = 0; i < a.length; i++)
      a[i] = aa.get(i);
  }

  static void sort(int[] a) {
    ArrayList<Integer> aa = new ArrayList<>();
    for (int i : a) {
      aa.add(i);
    }
    Collections.sort(aa);
    for (int i = 0; i < a.length; i++)
      a[i] = aa.get(i);
  }

  static long pow(long x, long y) {
    long res = 1l;
    while (y != 0) {
      if (y % 2 == 1) {
        res = x * res;
      }
      y /= 2;
      x = x * x;
    }
    return res;
  }

}

class dis {
  int x;
  int y;

  dis(int x, int y) {
    this.x = x;
    this.y = y;
  }
}