#include <bits/stdc++.h>
using namespace std;
string upper(string s) {
  transform(s.begin(), s.end(), s.begin(), ::toupper);
  return s;
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second > b.second);
}
long long modInverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
set<int> primeFactors(int n) {
  set<int> s;
  while (n % 2 == 0) {
    s.insert(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      s.insert(i);
      n = n / i;
    }
  }
  if (n > 2) s.insert(n);
  return s;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cout << 2 << "\n";
    int ans = n;
    int j = n - 1;
    for (int i = 0; i < n - 1; i++) {
      cout << j << " " << ans << "\n";
      ans = (ans + j + 1) / 2;
      j--;
    }
  }
  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;
string upper(string s) {
  transform(s.begin(), s.end(), s.begin(), ::toupper);
  return s;
}
bool isPrime(int n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (int i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
bool cmp(const pair<int, int> &a, const pair<int, int> &b) {
  return (a.second > b.second);
}
long long modInverse(long long a, long long m) {
  long long m0 = m;
  long long y = 0, x = 1;
  if (m == 1) return 0;
  while (a > 1) {
    long long q = a / m;
    long long t = m;
    m = a % m, a = t;
    t = y;
    y = x - q * y;
    x = t;
  }
  if (x < 0) x += m0;
  return x;
}
set<int> primeFactors(int n) {
  set<int> s;
  while (n % 2 == 0) {
    s.insert(2);
    n = n / 2;
  }
  for (int i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      s.insert(i);
      n = n / i;
    }
  }
  if (n > 2) s.insert(n);
  return s;
}
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cout << 2 << "\n";
    int ans = n;
    int j = n - 1;
    for (int i = 0; i < n - 1; i++) {
      cout << j << " " << ans << "\n";
      ans = (ans + j + 1) / 2;
      j--;
    }
  }
  return 0;
}

```