#include <bits/stdc++.h>
struct pairs {
  int a;
  int b;
};
int main() {
  using namespace std;
  int t, n, i, o, s;
  cin >> t;
  while (t--) {
    cin >> n;
    s = n;
    vector<int> a(n);
    vector<pairs> answer(n - 1);
    for (i = 1; i <= n; i++) {
      a[i - 1] = i;
    }
    for (o = 0; o < n - 1; o++) {
      for (i = s - 2; i >= 0; i--) {
        if (a[s - 1] % 2 == a[i] % 2) {
          answer[o].a = a[i];
          answer[o].b = a[s - 1];
          a.erase(a.begin() + i);
          s--;
          a.erase(a.begin() + s - 1);
          a.push_back((answer[o].a + answer[o].b) / 2);
          break;
        }
      }
    }
    cout << a[s - 1] << endl;
    if (n == 2)
      cout << 2 << ' ' << 1 << endl;
    else {
      for (pairs pair : answer) {
        cout << pair.a << ' ' << pair.b << endl;
      }
    }
  }
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

Numbers 1, 2, 3, ... n (each integer from 1 to n once) are written on a board. In one operation you can erase any two numbers a and b from the board and write one integer (a + b)/(2) rounded up instead.

You should perform the given operation n - 1 times and make the resulting number that will be left on the board as small as possible. 

For example, if n = 4, the following course of action is optimal:

  1. choose a = 4 and b = 2, so the new number is 3, and the whiteboard contains [1, 3, 3]; 
  2. choose a = 3 and b = 3, so the new number is 3, and the whiteboard contains [1, 3]; 
  3. choose a = 1 and b = 3, so the new number is 2, and the whiteboard contains [2]. 



It's easy to see that after n - 1 operations, there will be left only one number. Your goal is to minimize it.

Input

The first line contains one integer t (1 ≤ t ≤ 1000) — the number of test cases.

The only line of each test case contains one integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of integers written on the board initially.

It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, in the first line, print the minimum possible number left on the board after n - 1 operations. Each of the next n - 1 lines should contain two integers — numbers a and b chosen and erased in each operation.

Example

Input


1
4


Output


2
2 4
3 3
3 1

### Response

```cpp
#include <bits/stdc++.h>
struct pairs {
  int a;
  int b;
};
int main() {
  using namespace std;
  int t, n, i, o, s;
  cin >> t;
  while (t--) {
    cin >> n;
    s = n;
    vector<int> a(n);
    vector<pairs> answer(n - 1);
    for (i = 1; i <= n; i++) {
      a[i - 1] = i;
    }
    for (o = 0; o < n - 1; o++) {
      for (i = s - 2; i >= 0; i--) {
        if (a[s - 1] % 2 == a[i] % 2) {
          answer[o].a = a[i];
          answer[o].b = a[s - 1];
          a.erase(a.begin() + i);
          s--;
          a.erase(a.begin() + s - 1);
          a.push_back((answer[o].a + answer[o].b) / 2);
          break;
        }
      }
    }
    cout << a[s - 1] << endl;
    if (n == 2)
      cout << 2 << ' ' << 1 << endl;
    else {
      for (pairs pair : answer) {
        cout << pair.a << ' ' << pair.b << endl;
      }
    }
  }
  return 0;
}

```