#include <bits/stdc++.h>
using namespace std;
template <typename T>
T gcd(T a, T b) {
  while (a != 0 && b != 0) {
    if (a > b) {
      a %= b;
    } else {
      b %= a;
    }
  }
  return (a > b ? a : b);
}
long long exp(long long n, long long m, long long md = 1LL) {
  long long a = 1;
  while (m > 0) {
    if (m & 1) a = (a * n % md) % md;
    m /= 2;
    n = (n * n) % md;
  }
  return a % md;
}
void solve() {
  int n;
  cin >> n;
  multiset<int> m;
  vector<pair<int, int>> ans;
  for (int i = 1; i <= n; i++) m.insert(i);
  for (int i = 0; i < n - 1; i++) {
    auto it = m.end();
    it--;
    int a = *it;
    m.erase(it);
    it = m.end();
    it--;
    int b = *it;
    m.erase(it);
    m.insert((a + b + 1) / 2);
    ans.push_back(make_pair(a, b));
  }
  cout << *m.begin() << '\n';
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first << ' ' << ans[i].second << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

### Prompt

Your task is to create 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;
template <typename T>
T gcd(T a, T b) {
  while (a != 0 && b != 0) {
    if (a > b) {
      a %= b;
    } else {
      b %= a;
    }
  }
  return (a > b ? a : b);
}
long long exp(long long n, long long m, long long md = 1LL) {
  long long a = 1;
  while (m > 0) {
    if (m & 1) a = (a * n % md) % md;
    m /= 2;
    n = (n * n) % md;
  }
  return a % md;
}
void solve() {
  int n;
  cin >> n;
  multiset<int> m;
  vector<pair<int, int>> ans;
  for (int i = 1; i <= n; i++) m.insert(i);
  for (int i = 0; i < n - 1; i++) {
    auto it = m.end();
    it--;
    int a = *it;
    m.erase(it);
    it = m.end();
    it--;
    int b = *it;
    m.erase(it);
    m.insert((a + b + 1) / 2);
    ans.push_back(make_pair(a, b));
  }
  cout << *m.begin() << '\n';
  for (int i = 0; i < ans.size(); i++)
    cout << ans[i].first << ' ' << ans[i].second << '\n';
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
}

```