#include <bits/stdc++.h>
using namespace std;
template <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  ~debug() { cerr << endl; }
  template <class c>
  typename enable_if<sizeof dud<c>(0) != 1, debug&>::type operator<<(c i) {
    cerr << boolalpha << i;
    return *this;
  }
  template <class c>
  typename enable_if<sizeof dud<c>(0) == 1, debug&>::type operator<<(c i) {
    return *this << range(begin(i), end(i));
  }
  template <class c, class b>
  debug& operator<<(pair<b, c> d) {
    return *this << "(" << d.first << ", " << d.second << ")";
  }
  template <class c>
  debug& operator<<(rge<c> d) {
    *this << "[";
    for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
    return *this << "]";
  }
};
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    if (n == 2) {
      cout << "2" << endl;
      cout << "1 2" << endl;
    } else {
      int a[n];
      for (int i = 1; i <= n; i++) a[i] = i;
      cout << "2" << endl;
      cout << n << " " << n - 2 << endl;
      int sum = (n + (n - 2)) / 2;
      a[n - 2] = sum;
      for (int i = n - 1; i > 1; i--) {
        cout << a[i] << " " << a[i - 1] << endl;
        int dif = (a[i] + a[i - 1]) / 2;
        a[i - 1] = dif;
      }
    }
  }
}

### Prompt

Develop a solution in Cpp to the problem described below:

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 <class c>
struct rge {
  c b, e;
};
template <class c>
rge<c> range(c i, c j) {
  return rge<c>{i, j};
}
template <class c>
auto dud(c* x) -> decltype(cerr << *x, 0);
template <class c>
char dud(...);
struct debug {
  ~debug() { cerr << endl; }
  template <class c>
  typename enable_if<sizeof dud<c>(0) != 1, debug&>::type operator<<(c i) {
    cerr << boolalpha << i;
    return *this;
  }
  template <class c>
  typename enable_if<sizeof dud<c>(0) == 1, debug&>::type operator<<(c i) {
    return *this << range(begin(i), end(i));
  }
  template <class c, class b>
  debug& operator<<(pair<b, c> d) {
    return *this << "(" << d.first << ", " << d.second << ")";
  }
  template <class c>
  debug& operator<<(rge<c> d) {
    *this << "[";
    for (auto it = d.b; it != d.e; ++it) *this << ", " + 2 * (it == d.b) << *it;
    return *this << "]";
  }
};
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    int n;
    scanf("%d", &n);
    if (n == 2) {
      cout << "2" << endl;
      cout << "1 2" << endl;
    } else {
      int a[n];
      for (int i = 1; i <= n; i++) a[i] = i;
      cout << "2" << endl;
      cout << n << " " << n - 2 << endl;
      int sum = (n + (n - 2)) / 2;
      a[n - 2] = sum;
      for (int i = n - 1; i > 1; i--) {
        cout << a[i] << " " << a[i - 1] << endl;
        int dif = (a[i] + a[i - 1]) / 2;
        a[i - 1] = dif;
      }
    }
  }
}

```