#include <bits/stdc++.h>
using namespace std;
vector<char> alphabet = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                         'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                         's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
vector<char> calphabet = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
                          'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R',
                          'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, ans = 0;
    cin >> n;
    cout << "2\n" << n - 1 << " " << n << endl;
    for (long long it = n; it > 2; --it) cout << it - 2 << " " << it << endl;
  }
  return 0;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
vector<char> alphabet = {'a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i',
                         'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r',
                         's', 't', 'u', 'v', 'w', 'x', 'y', 'z'};
vector<char> calphabet = {'A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I',
                          'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R',
                          'S', 'T', 'U', 'V', 'W', 'X', 'Y', 'Z'};
int main() {
  long long t;
  cin >> t;
  while (t--) {
    long long n, ans = 0;
    cin >> n;
    cout << "2\n" << n - 1 << " " << n << endl;
    for (long long it = n; it > 2; --it) cout << it - 2 << " " << it << endl;
  }
  return 0;
}

```