#include <bits/stdc++.h>
using namespace std;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long hcf(long long n1, long long n2) {
  if (n2 != 0)
    return hcf(n2, n1 % n2);
  else
    return n1;
}
long long fact(long long n) {
  long long i = 1, co = 1;
  for (i = 1; i <= n; i++) co *= i;
  return co;
}
long long isprime(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (long long i = 5; i <= sqrt(n); i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  }
  return 1;
}
long long sod(long long n) {
  long long sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cout << 2 << '\n';
    long long q = n - 1, w = n;
    for (int i = 1; i <= n - 1; ++i) {
      cout << q << " " << w << '\n';
      w = ceil(((double)q + w) / 2);
      q--;
    }
  }
  return 0;
}

### 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;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
long long hcf(long long n1, long long n2) {
  if (n2 != 0)
    return hcf(n2, n1 % n2);
  else
    return n1;
}
long long fact(long long n) {
  long long i = 1, co = 1;
  for (i = 1; i <= n; i++) co *= i;
  return co;
}
long long isprime(long long n) {
  if (n <= 1) return 0;
  if (n <= 3) return 1;
  if (n % 2 == 0 || n % 3 == 0) return 0;
  for (long long i = 5; i <= sqrt(n); i += 6) {
    if (n % i == 0 || n % (i + 2) == 0) return 0;
  }
  return 1;
}
long long sod(long long n) {
  long long sum = 0;
  while (n) {
    sum += n % 10;
    n /= 10;
  }
  return sum;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t;
  t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cout << 2 << '\n';
    long long q = n - 1, w = n;
    for (int i = 1; i <= n - 1; ++i) {
      cout << q << " " << w << '\n';
      w = ceil(((double)q + w) / 2);
      q--;
    }
  }
  return 0;
}

```