#include <bits/stdc++.h>
const long long INF = 1e18;
const long long mINF = (-1) * INF;
const long long inf = 1e5 + 5;
using namespace std;
long long power(long long a, long long b) {
  long long result = 1;
  while (b) {
    if (b % 2) result = (result * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return result;
}
map<int, vector<int>> graph;
vector<int> visited(inf, 0);
vector<int> dist(inf);
void dfs(int node, int color) {
  visited[node] = 1;
  for (auto child : graph[node]) {
    if (visited[child] == 0) {
      dist[child] = 1 - color;
      dfs(child, 1 - color);
    }
  }
}
void solve() {
  int n, i, u, v;
  cin >> n;
  vector<int> a(n);
  for (i = 0; i < n; i++) {
    a[i] = i + 1;
  }
  cout << 2 << endl;
  u = n - 1;
  v = n - 2;
  for (i = 0; i < n - 1; i++) {
    cout << a[u] << " " << a[v] << endl;
    a[v] = (a[u] + a[v] + 1) / 2;
    u--;
    v--;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

### Prompt

Generate 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 long long INF = 1e18;
const long long mINF = (-1) * INF;
const long long inf = 1e5 + 5;
using namespace std;
long long power(long long a, long long b) {
  long long result = 1;
  while (b) {
    if (b % 2) result = (result * a) % 1000000007;
    a = (a * a) % 1000000007;
    b /= 2;
  }
  return result;
}
map<int, vector<int>> graph;
vector<int> visited(inf, 0);
vector<int> dist(inf);
void dfs(int node, int color) {
  visited[node] = 1;
  for (auto child : graph[node]) {
    if (visited[child] == 0) {
      dist[child] = 1 - color;
      dfs(child, 1 - color);
    }
  }
}
void solve() {
  int n, i, u, v;
  cin >> n;
  vector<int> a(n);
  for (i = 0; i < n; i++) {
    a[i] = i + 1;
  }
  cout << 2 << endl;
  u = n - 1;
  v = n - 2;
  for (i = 0; i < n - 1; i++) {
    cout << a[u] << " " << a[v] << endl;
    a[v] = (a[u] + a[v] + 1) / 2;
    u--;
    v--;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int t = 1;
  cin >> t;
  while (t--) {
    solve();
  }
  return 0;
}

```