#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    multiset<int, greater<int> > ms;
    std::vector<int> a, b;
    for (int i = 1; i <= n; i++) ms.insert(i);
    while (ms.size() > 2) {
      int arr[3];
      int cc = 0;
      for (auto x : ms) {
        arr[cc++] = x;
        if (cc > 1) break;
      }
      a.push_back(arr[0]), b.push_back(arr[1]), ms.erase(ms.find(arr[0])),
          ms.erase(ms.find(arr[1])), ms.insert((arr[0] + arr[1] + 1) / 2);
    }
    int arr[3];
    int cc = 0;
    for (auto x : ms) {
      arr[cc++] = x;
      if (cc > 1) break;
    }
    a.push_back(arr[0]);
    b.push_back(arr[1]);
    cout << (arr[0] + arr[1] + 1) / 2 << "\n";
    for (int i = 0; i < a.size(); i++) cout << a[i] << " " << b[i] << "\n";
  }
}

### Prompt

Your challenge is to write 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(0);
  cin.tie(0);
  int tc;
  cin >> tc;
  while (tc--) {
    int n;
    cin >> n;
    multiset<int, greater<int> > ms;
    std::vector<int> a, b;
    for (int i = 1; i <= n; i++) ms.insert(i);
    while (ms.size() > 2) {
      int arr[3];
      int cc = 0;
      for (auto x : ms) {
        arr[cc++] = x;
        if (cc > 1) break;
      }
      a.push_back(arr[0]), b.push_back(arr[1]), ms.erase(ms.find(arr[0])),
          ms.erase(ms.find(arr[1])), ms.insert((arr[0] + arr[1] + 1) / 2);
    }
    int arr[3];
    int cc = 0;
    for (auto x : ms) {
      arr[cc++] = x;
      if (cc > 1) break;
    }
    a.push_back(arr[0]);
    b.push_back(arr[1]);
    cout << (arr[0] + arr[1] + 1) / 2 << "\n";
    for (int i = 0; i < a.size(); i++) cout << a[i] << " " << b[i] << "\n";
  }
}

```