#include <bits/stdc++.h>
using namespace std;
inline int read() {
  int p = 0, w = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') p = p * 10 + c - '0', c = getchar();
  return p * w;
}
const int sid = 3e5 + 5;
int n;
int a[sid], b[sid];
priority_queue<int> q;
int main() {
  int t = read();
  while (t--) {
    n = read();
    for (register int i = 1; i <= n; i++) q.push(i);
    for (register int i = 1; i <= n - 1; i++) {
      a[i] = q.top();
      q.pop();
      b[i] = q.top();
      q.pop();
      q.push((a[i] + b[i] + 1) / 2);
    }
    printf("%d\n", q.top());
    q.pop();
    for (register int i = 1; i <= n - 1; i++) printf("%d %d\n", a[i], b[i]);
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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;
inline int read() {
  int p = 0, w = 1;
  char c = getchar();
  while (c > '9' || c < '0') {
    if (c == '-') w = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') p = p * 10 + c - '0', c = getchar();
  return p * w;
}
const int sid = 3e5 + 5;
int n;
int a[sid], b[sid];
priority_queue<int> q;
int main() {
  int t = read();
  while (t--) {
    n = read();
    for (register int i = 1; i <= n; i++) q.push(i);
    for (register int i = 1; i <= n - 1; i++) {
      a[i] = q.top();
      q.pop();
      b[i] = q.top();
      q.pop();
      q.push((a[i] + b[i] + 1) / 2);
    }
    printf("%d\n", q.top());
    q.pop();
    for (register int i = 1; i <= n - 1; i++) printf("%d %d\n", a[i], b[i]);
  }
  return 0;
}

```