#include <bits/stdc++.h>
using namespace std;
void fun2(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
void fun(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) cin >> a[i];
}
void vec(vector<long long int> &v, long long int n) {
  for (long long int i = 0; i < n; i++) {
    long long int x, z;
    cin >> x;
    v.push_back(x);
  }
}
void vec2(vector<long long int> &v) {
  for (auto i = v.begin(); i != v.end(); ++i) cout << *i << " ";
  cout << "\n";
}
void vecr(vector<long long int> &v) {
  for (auto i = v.rbegin(); i != v.rend(); ++i) cout << *i << " ";
  cout << "\n";
}
void set1(set<long long int> &s, long long int &n) {
  for (long long int i = 0; i < n; i++) {
    long long int x, z;
    cin >> x;
    s.insert(x);
  }
}
void set2(set<long long int> &s) {
  for (auto i = s.begin(); i != s.end(); ++i) cout << *i << " ";
  cout << "\n";
}
long long int max(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
long long int min(long long int a, long long int b) {
  if (a > b)
    return b;
  else
    return a;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, s = 0;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; i++) {
      a[i] = i + 1;
    }
    cout << 2;
    cout << "\n";
    for (long long int i = 0; i < n - 1; i++) {
      cout << a[n - 1 - i] << " " << a[n - 2 - i];
      a[n - 2 - i] = ceil((a[n - 1 - i] + a[n - 2 - i]) * 0.5);
      cout << "\n";
    }
  }
  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;
void fun2(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) cout << a[i] << " ";
  cout << "\n";
}
void fun(long long int a[], long long int n) {
  for (long long int i = 0; i < n; i++) cin >> a[i];
}
void vec(vector<long long int> &v, long long int n) {
  for (long long int i = 0; i < n; i++) {
    long long int x, z;
    cin >> x;
    v.push_back(x);
  }
}
void vec2(vector<long long int> &v) {
  for (auto i = v.begin(); i != v.end(); ++i) cout << *i << " ";
  cout << "\n";
}
void vecr(vector<long long int> &v) {
  for (auto i = v.rbegin(); i != v.rend(); ++i) cout << *i << " ";
  cout << "\n";
}
void set1(set<long long int> &s, long long int &n) {
  for (long long int i = 0; i < n; i++) {
    long long int x, z;
    cin >> x;
    s.insert(x);
  }
}
void set2(set<long long int> &s) {
  for (auto i = s.begin(); i != s.end(); ++i) cout << *i << " ";
  cout << "\n";
}
long long int max(long long int a, long long int b) {
  if (a > b)
    return a;
  else
    return b;
}
long long int min(long long int a, long long int b) {
  if (a > b)
    return b;
  else
    return a;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n, s = 0;
    cin >> n;
    long long int a[n];
    for (long long int i = 0; i < n; i++) {
      a[i] = i + 1;
    }
    cout << 2;
    cout << "\n";
    for (long long int i = 0; i < n - 1; i++) {
      cout << a[n - 1 - i] << " " << a[n - 2 - i];
      a[n - 2 - i] = ceil((a[n - 1 - i] + a[n - 2 - i]) * 0.5);
      cout << "\n";
    }
  }
  return 0;
}

```