#include <bits/stdc++.h>
using namespace std;
int const MAXN = 2e6 + 9;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    multiset<int> odd, even;
    vector<pair<int, int>> out;
    for (int i = 2; i <= n; i++) {
      if (i & 1)
        odd.insert(i);
      else
        even.insert(i);
    }
    int ans = 0;
    for (int i = 0; (int)out.size() < n - 2; i++) {
      int add;
      bool goEven = false, goOdd = false;
      if ((int)even.size() >= 2 && (int)odd.size() >= 2) {
        if (*even.rbegin() >= *odd.rbegin()) {
          goEven = true;
        } else
          goOdd = true;
      } else if ((int)even.size() >= 2) {
        goEven = true;
      } else if ((int)odd.size() >= 2) {
        goOdd = true;
      }
      if (goEven) {
        auto mx = even.find(*even.rbegin());
        auto mn = even.find(*even.rbegin());
        mn--;
        out.emplace_back(*mx, *mn);
        add = (*mn + *mx + 1) / 2;
        even.erase(mn);
        even.erase(mx);
      } else if (goOdd) {
        auto mx = odd.find(*odd.rbegin());
        auto mn = odd.find(*odd.rbegin());
        mn--;
        out.emplace_back(*mx, *mn);
        add = (*mn + *mx + 1) / 2;
        odd.erase(mn);
        odd.erase(mx);
      } else {
        int mn = *even.begin();
        int mx = *odd.begin();
        out.emplace_back(mx, mn);
        even.erase(even.begin());
        odd.erase(odd.begin());
        add = (mn + mx + 1) / 2;
      }
      if (add & 1) {
        if (odd.count(add) == 1) {
          out.emplace_back(add, add);
        } else
          odd.insert(add);
      } else {
        if (even.count(add) == 1) {
          out.emplace_back(add, add);
        } else
          even.insert(add);
      }
    }
    int x = (odd.size()) ? *odd.begin() : *even.begin();
    out.emplace_back(x, 1);
    ans = (x + 2) / 2;
    cout << ans << '\n';
    for (auto cur : out) {
      cout << cur.first << " " << cur.second << '\n';
    }
  }
}

### Prompt

Construct a cpp code solution to the problem outlined:

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;
int const MAXN = 2e6 + 9;
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    multiset<int> odd, even;
    vector<pair<int, int>> out;
    for (int i = 2; i <= n; i++) {
      if (i & 1)
        odd.insert(i);
      else
        even.insert(i);
    }
    int ans = 0;
    for (int i = 0; (int)out.size() < n - 2; i++) {
      int add;
      bool goEven = false, goOdd = false;
      if ((int)even.size() >= 2 && (int)odd.size() >= 2) {
        if (*even.rbegin() >= *odd.rbegin()) {
          goEven = true;
        } else
          goOdd = true;
      } else if ((int)even.size() >= 2) {
        goEven = true;
      } else if ((int)odd.size() >= 2) {
        goOdd = true;
      }
      if (goEven) {
        auto mx = even.find(*even.rbegin());
        auto mn = even.find(*even.rbegin());
        mn--;
        out.emplace_back(*mx, *mn);
        add = (*mn + *mx + 1) / 2;
        even.erase(mn);
        even.erase(mx);
      } else if (goOdd) {
        auto mx = odd.find(*odd.rbegin());
        auto mn = odd.find(*odd.rbegin());
        mn--;
        out.emplace_back(*mx, *mn);
        add = (*mn + *mx + 1) / 2;
        odd.erase(mn);
        odd.erase(mx);
      } else {
        int mn = *even.begin();
        int mx = *odd.begin();
        out.emplace_back(mx, mn);
        even.erase(even.begin());
        odd.erase(odd.begin());
        add = (mn + mx + 1) / 2;
      }
      if (add & 1) {
        if (odd.count(add) == 1) {
          out.emplace_back(add, add);
        } else
          odd.insert(add);
      } else {
        if (even.count(add) == 1) {
          out.emplace_back(add, add);
        } else
          even.insert(add);
      }
    }
    int x = (odd.size()) ? *odd.begin() : *even.begin();
    out.emplace_back(x, 1);
    ans = (x + 2) / 2;
    cout << ans << '\n';
    for (auto cur : out) {
      cout << cur.first << " " << cur.second << '\n';
    }
  }
}

```