#include <bits/stdc++.h>
using namespace std;
void solve() {
  int n;
  cin >> n;
  priority_queue<int> pq{};
  for (int i{1}; i <= n; ++i) pq.push(i);
  vector<pair<int, int>> ans{};
  for (int i{0}; i < n - 1; ++i) {
    int mx1{pq.top()};
    pq.pop();
    int mx2{pq.top()};
    pq.pop();
    ans.push_back({mx1, mx2});
    pq.push((mx1 + mx2 + 1) / 2);
  }
  cout << pq.top() << '\n';
  for (auto& x : ans) cout << x.first << " " << x.second << '\n';
}
void solve2() {
  int n;
  cin >> n;
  cout << "2\n";
  cout << n << ' ' << n - 1 << '\n';
  for (int i{n}; i >= 3; --i) cout << i << ' ' << i - 2 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve2();
  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>
using namespace std;
void solve() {
  int n;
  cin >> n;
  priority_queue<int> pq{};
  for (int i{1}; i <= n; ++i) pq.push(i);
  vector<pair<int, int>> ans{};
  for (int i{0}; i < n - 1; ++i) {
    int mx1{pq.top()};
    pq.pop();
    int mx2{pq.top()};
    pq.pop();
    ans.push_back({mx1, mx2});
    pq.push((mx1 + mx2 + 1) / 2);
  }
  cout << pq.top() << '\n';
  for (auto& x : ans) cout << x.first << " " << x.second << '\n';
}
void solve2() {
  int n;
  cin >> n;
  cout << "2\n";
  cout << n << ' ' << n - 1 << '\n';
  for (int i{n}; i >= 3; --i) cout << i << ' ' << i - 2 << '\n';
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  int t;
  cin >> t;
  while (t--) solve2();
  return 0;
}

```