#include <bits/stdc++.h>
using namespace std;
namespace io {
inline int rdi() {
  int v;
  cin >> v;
  return v;
}
inline long long rdll() {
  long long v;
  cin >> v;
  return v;
}
inline string rds() {
  string s;
  cin >> s;
  return s;
}
inline char rdc() {
  char c;
  cin >> c;
  return c;
}
template <class T>
void rdv(T beg_iter, T end_iter) {
  for (; beg_iter != end_iter; ++beg_iter) cin >> *beg_iter;
}
inline void wri(const int v) { cout << v; }
inline void wrll(const long long& v) { cout << v; }
inline void wrs(const string& s) { cout << s; }
inline void wrc(const char c) { cout << c; }
template <class T>
void wrv(T beg_iter, T end_iter, string separator = " ",
         string end_of_line = "\n") {
  for (; beg_iter != end_iter; ++beg_iter) {
    cout << *beg_iter;
    cout << separator;
  }
  cout << end_of_line;
}
}  // namespace io
using namespace io;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  if (10 == 1) {
    t = 1;
  } else {
    cin >> t;
  }
  while (t--) solve();
  return 0;
}
void solve() {
  long long n;
  cin >> n;
  cout << 2 << '\n';
  int l = n - 1, r = n;
  for (long long i = 0; i < n - 1; ++i) {
    cout << l << " " << r << '\n';
    r = (l + r + 1) / 2;
    l--;
  }
}

### Prompt

In cpp, your task is to solve 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;
namespace io {
inline int rdi() {
  int v;
  cin >> v;
  return v;
}
inline long long rdll() {
  long long v;
  cin >> v;
  return v;
}
inline string rds() {
  string s;
  cin >> s;
  return s;
}
inline char rdc() {
  char c;
  cin >> c;
  return c;
}
template <class T>
void rdv(T beg_iter, T end_iter) {
  for (; beg_iter != end_iter; ++beg_iter) cin >> *beg_iter;
}
inline void wri(const int v) { cout << v; }
inline void wrll(const long long& v) { cout << v; }
inline void wrs(const string& s) { cout << s; }
inline void wrc(const char c) { cout << c; }
template <class T>
void wrv(T beg_iter, T end_iter, string separator = " ",
         string end_of_line = "\n") {
  for (; beg_iter != end_iter; ++beg_iter) {
    cout << *beg_iter;
    cout << separator;
  }
  cout << end_of_line;
}
}  // namespace io
using namespace io;
void solve();
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  ;
  long long t;
  if (10 == 1) {
    t = 1;
  } else {
    cin >> t;
  }
  while (t--) solve();
  return 0;
}
void solve() {
  long long n;
  cin >> n;
  cout << 2 << '\n';
  int l = n - 1, r = n;
  for (long long i = 0; i < n - 1; ++i) {
    cout << l << " " << r << '\n';
    r = (l + r + 1) / 2;
    l--;
  }
}

```