#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    long long int n, i, j, k;
    cin >> n;
    priority_queue<int> pq;
    vector<pair<int, int>> v;
    for (i = 0; i < n; i++) pq.push(i + 1);
    while (pq.size() >= 3) {
      int k1 = pq.top();
      pq.pop();
      int k2 = pq.top();
      pq.pop();
      if ((k1 + k2) % 2 == 0) {
        v.push_back(make_pair(k1, k2));
        pq.push((k1 + k2) / 2);
      } else {
        int k3 = pq.top();
        pq.pop();
        pq.push(k2);
        v.push_back(make_pair(k1, k3));
        int avg = (int)ceil((k1 + k3) / 2.0);
        pq.push(avg);
      }
    }
    int z1 = pq.top();
    pq.pop();
    int z2 = pq.top();
    pq.pop();
    v.push_back(make_pair(z1, z2));
    cout << ceil((z1 + z2) / 2.0) << "\n";
    for (i = 0; i < v.size(); i++)
      cout << v[i].first << " " << v[i].second << "\n";
  }
  return 0;
}

### Prompt

Please provide a CPP coded solution 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>
using namespace std;
const int mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  ;
  int T;
  cin >> T;
  while (T--) {
    long long int n, i, j, k;
    cin >> n;
    priority_queue<int> pq;
    vector<pair<int, int>> v;
    for (i = 0; i < n; i++) pq.push(i + 1);
    while (pq.size() >= 3) {
      int k1 = pq.top();
      pq.pop();
      int k2 = pq.top();
      pq.pop();
      if ((k1 + k2) % 2 == 0) {
        v.push_back(make_pair(k1, k2));
        pq.push((k1 + k2) / 2);
      } else {
        int k3 = pq.top();
        pq.pop();
        pq.push(k2);
        v.push_back(make_pair(k1, k3));
        int avg = (int)ceil((k1 + k3) / 2.0);
        pq.push(avg);
      }
    }
    int z1 = pq.top();
    pq.pop();
    int z2 = pq.top();
    pq.pop();
    v.push_back(make_pair(z1, z2));
    cout << ceil((z1 + z2) / 2.0) << "\n";
    for (i = 0; i < v.size(); i++)
      cout << v[i].first << " " << v[i].second << "\n";
  }
  return 0;
}

```