#include <bits/stdc++.h>
using namespace std;
template <typename T>
inline void read(T& t) {
  t = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    t = t * 10 + ch - '0';
    ch = getchar();
  }
  t *= f;
}
void TEST() { cout << "OK!!!" << endl; }
const int dx[4] = {1, -1, 1, -1};
const int dy[4] = {-1, 1, 1, -1};
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-5;
const int maxn = 2e5 + 100;
int n, m;
priority_queue<int> q;
int main() {
  int t;
  read(t);
  while (t--) {
    read(n);
    for (int i = 1; i <= n; i++) q.push(i);
    cout << 2 << endl;
    while (q.size() > 1) {
      int a = q.top();
      q.pop();
      int b = q.top();
      q.pop();
      int tmp = (a + b) * 1.0 / 2 + 0.5;
      q.push(tmp);
      cout << a << " " << b << endl;
    }
    q.pop();
  }
  return 0;
}

### Prompt

Generate 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;
template <typename T>
inline void read(T& t) {
  t = 0;
  int f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (isdigit(ch)) {
    t = t * 10 + ch - '0';
    ch = getchar();
  }
  t *= f;
}
void TEST() { cout << "OK!!!" << endl; }
const int dx[4] = {1, -1, 1, -1};
const int dy[4] = {-1, 1, 1, -1};
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-5;
const int maxn = 2e5 + 100;
int n, m;
priority_queue<int> q;
int main() {
  int t;
  read(t);
  while (t--) {
    read(n);
    for (int i = 1; i <= n; i++) q.push(i);
    cout << 2 << endl;
    while (q.size() > 1) {
      int a = q.top();
      q.pop();
      int b = q.top();
      q.pop();
      int tmp = (a + b) * 1.0 / 2 + 0.5;
      q.push(tmp);
      cout << a << " " << b << endl;
    }
    q.pop();
  }
  return 0;
}

```