#include <bits/stdc++.h>
using namespace std;
const long long mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long n, i;
    cin >> n;
    if (n == 2) {
      cout << 2 << "\n";
      cout << "1 2";
      cout << endl;
      continue;
      ;
    }
    vector<long long> a(n, 0);
    for (long long i = 0; i < n; i++) {
      a[i] = i + 1;
    }
    reverse(a.begin(), a.end());
    pair<long long, long long> p[n - 1];
    for (i = 0; i < n - 1; i++) {
      long long x = a[i], y = a[i + 1];
      if ((x + y) % 2) {
        a[i + 1] = (x + y) / 2 + 1;
      } else {
        a[i + 1] = (x + y) / 2;
      }
      a[i] = 0;
      p[i].first = x, p[i].second = y;
    }
    cout << a[n - 1];
    cout << endl;
    ;
    for (i = 0; i < n - 1; i++) {
      cout << p[i].first << " " << p[i].second;
      cout << endl;
      ;
    }
  }
}

### 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;
const long long mod = 1000000007;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  int t;
  cin >> t;
  while (t--) {
    long long n, i;
    cin >> n;
    if (n == 2) {
      cout << 2 << "\n";
      cout << "1 2";
      cout << endl;
      continue;
      ;
    }
    vector<long long> a(n, 0);
    for (long long i = 0; i < n; i++) {
      a[i] = i + 1;
    }
    reverse(a.begin(), a.end());
    pair<long long, long long> p[n - 1];
    for (i = 0; i < n - 1; i++) {
      long long x = a[i], y = a[i + 1];
      if ((x + y) % 2) {
        a[i + 1] = (x + y) / 2 + 1;
      } else {
        a[i + 1] = (x + y) / 2;
      }
      a[i] = 0;
      p[i].first = x, p[i].second = y;
    }
    cout << a[n - 1];
    cout << endl;
    ;
    for (i = 0; i < n - 1; i++) {
      cout << p[i].first << " " << p[i].second;
      cout << endl;
      ;
    }
  }
}

```