#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    for (int i = 0; i < n; i++) {
      arr[i] = i + 1;
    }
    cout << 2 << endl;
    for (int i = n - 2; i >= 0; i--) {
      double sum = (arr[i] + arr[i + 1]) / 2.0;
      int int_sum = (arr[i] + arr[i - 1]) / 2;
      if (sum - int_sum >= 0.5) {
        cout << arr[i] << " " << arr[i + 1] << endl;
        arr[i] = ceil(sum);
      } else {
        cout << arr[i] << " " << arr[i + 1] << endl;
        arr[i] = int_sum;
      }
    }
  }
}

### Prompt

Please formulate a CPP solution to the following problem:

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>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long arr[n];
    for (int i = 0; i < n; i++) {
      arr[i] = i + 1;
    }
    cout << 2 << endl;
    for (int i = n - 2; i >= 0; i--) {
      double sum = (arr[i] + arr[i + 1]) / 2.0;
      int int_sum = (arr[i] + arr[i - 1]) / 2;
      if (sum - int_sum >= 0.5) {
        cout << arr[i] << " " << arr[i + 1] << endl;
        arr[i] = ceil(sum);
      } else {
        cout << arr[i] << " " << arr[i + 1] << endl;
        arr[i] = int_sum;
      }
    }
  }
}

```