#include <bits/stdc++.h>
const int mod = 1000000007;
using namespace std;
void solve(vector<vector<int>> &arr, int u, int x, int y, int t, int cur,
           int &ans, unordered_map<string, int> &mp) {
  if (u == arr.size()) {
    ans = max(ans, cur);
    return;
  }
  string temp = to_string(u) + "-" + to_string(x) + "-" + to_string(y);
  if (mp.find(temp) != mp.end()) return;
  mp[temp] = 1;
  int time_left = arr[u][0] - t;
  int time_we_have = abs(arr[u][1] - x) + abs(arr[u][2] - y);
  if (time_left >= time_we_have)
    solve(arr, u + 1, arr[u][1], arr[u][2], arr[u][0], cur + 1, ans, mp);
  solve(arr, u + 1, x, y, t, cur, ans, mp);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  auto start = chrono::steady_clock::now();
  int TESTCASE = 1;
  cin >> TESTCASE;
  while (TESTCASE--) {
    int n;
    cin >> n;
    cout << "2\n";
    priority_queue<int> pq;
    for (int i = 1; i <= n; ++i) pq.push(i);
    while (pq.size() > 1) {
      int a = pq.top();
      pq.pop();
      int b = pq.top();
      pq.pop();
      cout << a << " " << b << endl;
      if ((a + b) % 2 == 1)
        pq.push((a + b + 1) / 2);
      else
        pq.push((a + b) / 2);
    }
  }
  auto end = chrono::steady_clock::now();
  return 0;
}

### Prompt

Your challenge is to write 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>
const int mod = 1000000007;
using namespace std;
void solve(vector<vector<int>> &arr, int u, int x, int y, int t, int cur,
           int &ans, unordered_map<string, int> &mp) {
  if (u == arr.size()) {
    ans = max(ans, cur);
    return;
  }
  string temp = to_string(u) + "-" + to_string(x) + "-" + to_string(y);
  if (mp.find(temp) != mp.end()) return;
  mp[temp] = 1;
  int time_left = arr[u][0] - t;
  int time_we_have = abs(arr[u][1] - x) + abs(arr[u][2] - y);
  if (time_left >= time_we_have)
    solve(arr, u + 1, arr[u][1], arr[u][2], arr[u][0], cur + 1, ans, mp);
  solve(arr, u + 1, x, y, t, cur, ans, mp);
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  auto start = chrono::steady_clock::now();
  int TESTCASE = 1;
  cin >> TESTCASE;
  while (TESTCASE--) {
    int n;
    cin >> n;
    cout << "2\n";
    priority_queue<int> pq;
    for (int i = 1; i <= n; ++i) pq.push(i);
    while (pq.size() > 1) {
      int a = pq.top();
      pq.pop();
      int b = pq.top();
      pq.pop();
      cout << a << " " << b << endl;
      if ((a + b) % 2 == 1)
        pq.push((a + b + 1) / 2);
      else
        pq.push((a + b) / 2);
    }
  }
  auto end = chrono::steady_clock::now();
  return 0;
}

```