#include <bits/stdc++.h>
using namespace std;
const long long mx = 10000007;
using pii = pair<long long, long long>;
const long long mod = 1e9 + 7;
const double pi = acos(-1);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<pii> v;
    vector<long long> ev, od;
    set<long long> s;
    for (long long i = 0; i < n; i++) {
      s.insert(i + 1);
      if ((i + 1) % 2 == 0)
        ev.push_back(i + 1);
      else
        od.push_back(i + 1);
    }
    if (n == 2) {
      cout << 2 << endl;
      cout << 1 << " " << 2 << endl;
      continue;
    }
    sort(ev.begin(), ev.end());
    sort(od.begin(), od.end());
    reverse(ev.begin(), ev.end());
    reverse(od.begin(), od.end());
    long long d;
    if (n % 2 == 0) {
      d = ev[0];
      s.erase(d);
      long long i = 1, j = 0;
      while (1) {
        if (d % 2 == 0) {
          long long a = d;
          long long b = ev[i];
          s.erase(b);
          v.push_back({a, b});
          d = (a + b) / 2;
          if (s.size() == 0) break;
          i++;
          if (i > ev.size()) break;
        } else {
          long long a = d;
          long long b = od[j];
          s.erase(b);
          v.push_back({a, b});
          j++;
          d = (a + b) / 2;
          if (s.size() == 0) break;
          if (j > od.size()) break;
        }
        if (s.size() == 0) break;
      }
    } else if (n % 2 == 1) {
      d = od[0];
      s.erase(d);
      long long i = 1, j = 0;
      while (1) {
        if (d % 2 == 1) {
          long long a = d;
          long long b = od[i];
          s.erase(b);
          v.push_back({a, b});
          d = (a + b) / 2;
          if (s.size() == 0) break;
          i++;
          if (i > od.size()) break;
        } else {
          long long a = d;
          long long b = ev[j];
          s.erase(b);
          v.push_back({a, b});
          j++;
          d = (a + b) / 2;
          if (s.size() == 0) break;
          if (j > ev.size()) break;
        }
        if (s.size() == 0) break;
      }
    }
    cout << d << endl;
    for (long long i = 0; i < v.size(); i++) {
      cout << v[i].first << " " << v[i].second << endl;
    }
  }
  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;
const long long mx = 10000007;
using pii = pair<long long, long long>;
const long long mod = 1e9 + 7;
const double pi = acos(-1);
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  ;
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    vector<pii> v;
    vector<long long> ev, od;
    set<long long> s;
    for (long long i = 0; i < n; i++) {
      s.insert(i + 1);
      if ((i + 1) % 2 == 0)
        ev.push_back(i + 1);
      else
        od.push_back(i + 1);
    }
    if (n == 2) {
      cout << 2 << endl;
      cout << 1 << " " << 2 << endl;
      continue;
    }
    sort(ev.begin(), ev.end());
    sort(od.begin(), od.end());
    reverse(ev.begin(), ev.end());
    reverse(od.begin(), od.end());
    long long d;
    if (n % 2 == 0) {
      d = ev[0];
      s.erase(d);
      long long i = 1, j = 0;
      while (1) {
        if (d % 2 == 0) {
          long long a = d;
          long long b = ev[i];
          s.erase(b);
          v.push_back({a, b});
          d = (a + b) / 2;
          if (s.size() == 0) break;
          i++;
          if (i > ev.size()) break;
        } else {
          long long a = d;
          long long b = od[j];
          s.erase(b);
          v.push_back({a, b});
          j++;
          d = (a + b) / 2;
          if (s.size() == 0) break;
          if (j > od.size()) break;
        }
        if (s.size() == 0) break;
      }
    } else if (n % 2 == 1) {
      d = od[0];
      s.erase(d);
      long long i = 1, j = 0;
      while (1) {
        if (d % 2 == 1) {
          long long a = d;
          long long b = od[i];
          s.erase(b);
          v.push_back({a, b});
          d = (a + b) / 2;
          if (s.size() == 0) break;
          i++;
          if (i > od.size()) break;
        } else {
          long long a = d;
          long long b = ev[j];
          s.erase(b);
          v.push_back({a, b});
          j++;
          d = (a + b) / 2;
          if (s.size() == 0) break;
          if (j > ev.size()) break;
        }
        if (s.size() == 0) break;
      }
    }
    cout << d << endl;
    for (long long i = 0; i < v.size(); i++) {
      cout << v[i].first << " " << v[i].second << endl;
    }
  }
  return 0;
}

```