#include <bits/stdc++.h>
using namespace std;
const long long MAXN = (long long)((1e5) + 100);
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
void prl(long long x) { cout << x << "\n"; }
const long long N = (long long)(1 * 1e6 + 10);
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) a[i] = i + 1;
    sort(a, a + n, greater<long long>());
    long long ans = (a[0] + a[1] + 1) / 2;
    for (long long i = 2; i < n; i++) {
      ans = (ans + a[i] + 1) / 2;
    }
    prl(ans);
    cout << a[0] << " " << a[1] << "\n";
    ans = (a[0] + a[1] + 1) / 2;
    for (long long i = 2; i < n; i++) {
      cout << a[i] << " " << ans << "\n";
      ans = (ans + a[i] + 1) / 2;
    }
  }
}

### 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;
const long long MAXN = (long long)((1e5) + 100);
long long gcd(long long a, long long b) {
  if (b == 0)
    return a;
  else
    return gcd(b, a % b);
}
long long max(long long a, long long b) {
  if (a > b)
    return a;
  else
    return b;
}
long long min(long long a, long long b) {
  if (a < b)
    return a;
  else
    return b;
}
void prl(long long x) { cout << x << "\n"; }
const long long N = (long long)(1 * 1e6 + 10);
using namespace std;
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    for (long long i = 0; i < n; i++) a[i] = i + 1;
    sort(a, a + n, greater<long long>());
    long long ans = (a[0] + a[1] + 1) / 2;
    for (long long i = 2; i < n; i++) {
      ans = (ans + a[i] + 1) / 2;
    }
    prl(ans);
    cout << a[0] << " " << a[1] << "\n";
    ans = (a[0] + a[1] + 1) / 2;
    for (long long i = 2; i < n; i++) {
      cout << a[i] << " " << ans << "\n";
      ans = (ans + a[i] + 1) / 2;
    }
  }
}

```