#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vs = vector<string>;
using vl = vector<ll>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;
using vvc = vector<vector<char>>;
using pii = pair<int, int>;
template <typename T>
void view(T e) {
  std::cout << e << std::endl;
}
template <typename T>
void view(const std::vector<T>& v) {
  for (const auto& e : v) {
    std::cout << e << " ";
  }
  std::cout << std::endl;
}
template <typename T>
void view(const std::vector<std::vector<T>>& vv) {
  cout << endl;
  int cnt = 0;
  for (const auto& v : vv) {
    cout << cnt << "th : ";
    view(v);
    cnt++;
  }
  cout << endl;
}
const ll mod = 1000000007;
const int inf = 1001001001;
const ll INF = 1001001001001001001;
const int MAX = 10000;
template <class T>
bool chmax(T& a, const T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
ll modpow(ll a, ll p) {
  ll ret = 1;
  while (p) {
    if (p & 1) {
      ret = ret * a % mod;
    }
    a = a * a % mod;
    p >>= 1;
  }
  return ret;
}
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cout << 2 << endl;
    cout << n << " " << n - 1 << endl;
    for (int i = n; i - 2 >= 1; i--) {
      cout << i << " " << i - 2 << endl;
    }
  }
}

### Prompt

Please create a solution in Cpp 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>
using namespace std;
using ll = long long;
using vi = vector<int>;
using vs = vector<string>;
using vl = vector<ll>;
using vb = vector<bool>;
using vvi = vector<vector<int>>;
using vvl = vector<vector<ll>>;
using vvc = vector<vector<char>>;
using pii = pair<int, int>;
template <typename T>
void view(T e) {
  std::cout << e << std::endl;
}
template <typename T>
void view(const std::vector<T>& v) {
  for (const auto& e : v) {
    std::cout << e << " ";
  }
  std::cout << std::endl;
}
template <typename T>
void view(const std::vector<std::vector<T>>& vv) {
  cout << endl;
  int cnt = 0;
  for (const auto& v : vv) {
    cout << cnt << "th : ";
    view(v);
    cnt++;
  }
  cout << endl;
}
const ll mod = 1000000007;
const int inf = 1001001001;
const ll INF = 1001001001001001001;
const int MAX = 10000;
template <class T>
bool chmax(T& a, const T b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <class T>
bool chmin(T& a, const T b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
ll modpow(ll a, ll p) {
  ll ret = 1;
  while (p) {
    if (p & 1) {
      ret = ret * a % mod;
    }
    a = a * a % mod;
    p >>= 1;
  }
  return ret;
}
int dx[4] = {1, 0, -1, 0};
int dy[4] = {0, 1, 0, -1};
int main() {
  int t;
  cin >> t;
  while (t--) {
    int n;
    cin >> n;
    cout << 2 << endl;
    cout << n << " " << n - 1 << endl;
    for (int i = n; i - 2 >= 1; i--) {
      cout << i << " " << i - 2 << endl;
    }
  }
}

```