#include <bits/stdc++.h>
using namespace std;
void fastIO() {
  ios_base::sync_with_stdio(false);
  cout.setf(ios::fixed);
  cout.precision(10);
  cout.tie(nullptr);
  cin.tie(nullptr);
}
template <class T>
T power(T N, T P) {
  return (P == 0) ? 1 : N * power(N, P - 1);
}
template <class T>
T GCD(T A, T B) {
  return (A % B == 0) ? B : GCD(B, A % B);
}
const int MOD = 1e9 + 7;
const int MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = acos((long double)-1);
void setIn(string s) { freopen(s.c_str(), "r", stdin); }
void setOut(string s) { freopen(s.c_str(), "w", stdout); }
bool sort_rev(int a, int b) { return a > b; }
bool sort_(pair<string, int> &p1, pair<string, int> &p2) {
  if (p1.first == p2.first) return p1.second < p2.second;
  return p1.first < p2.first;
}
int main() {
  fastIO();
  int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cout << 2 << "\n";
    int avg = n;
    for (int i = n - 1; i >= 1; i--) {
      cout << i << " " << avg << "\n";
      avg = (avg + i + 1) / 2;
    }
  }
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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;
void fastIO() {
  ios_base::sync_with_stdio(false);
  cout.setf(ios::fixed);
  cout.precision(10);
  cout.tie(nullptr);
  cin.tie(nullptr);
}
template <class T>
T power(T N, T P) {
  return (P == 0) ? 1 : N * power(N, P - 1);
}
template <class T>
T GCD(T A, T B) {
  return (A % B == 0) ? B : GCD(B, A % B);
}
const int MOD = 1e9 + 7;
const int MX = 2e5 + 5;
const long long INF = 1e18;
const long double PI = acos((long double)-1);
void setIn(string s) { freopen(s.c_str(), "r", stdin); }
void setOut(string s) { freopen(s.c_str(), "w", stdout); }
bool sort_rev(int a, int b) { return a > b; }
bool sort_(pair<string, int> &p1, pair<string, int> &p2) {
  if (p1.first == p2.first) return p1.second < p2.second;
  return p1.first < p2.first;
}
int main() {
  fastIO();
  int t = 1;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cout << 2 << "\n";
    int avg = n;
    for (int i = n - 1; i >= 1; i--) {
      cout << i << " " << avg << "\n";
      avg = (avg + i + 1) / 2;
    }
  }
  return 0;
}

```