#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 {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long int INF = 1e18;
void init() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool checkp2(long long int n) { return n && !(n & (n - 1)); }
long long int numsetbit(long long int n) {
  long long int count = 0;
  for (; n; count++) {
    n = n & (n - 1);
  }
  return count;
}
long long int powe(long long int x, long long int n, const long long int mod) {
  x = x % mod;
  long long int po = 1;
  while (n) {
    if (n & 1) {
      po = (po * x) % mod;
    }
    n = n >> 1;
    x = ((x % mod) * (x % mod)) % mod;
  }
  return po;
}
long long int powm(long long int x, long long int n) {
  long long int po = 1;
  while (n) {
    if (n & 1) {
      po = (po * x);
    }
    n = n >> 1;
    x = (x * x);
  }
  return po;
}
bool isPrime(long long int x) {
  if (x < 2) return 0;
  for (long long int y = 2; y * y <= x; y++) {
    if (x % y == 0) {
      return 0;
    }
  }
  return true;
}
const long long int mod = 1e9 + 7;
void solve() {
  long long int n;
  cin >> n;
  if (n == 2) {
    cout << "2" << endl;
    cout << "1 2" << endl;
    return;
  }
  cout << "2" << endl;
  cout << n << " " << n - 2 << endl;
  cout << n - 1 << " " << n - 1 << endl;
  for (long long int i = 0; i < n - 3; i++) {
    cout << n - i - 1 << " " << n - i - 3 << endl;
  }
}
int main() {
  init();
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
}

### 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>
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 {
  template <class c>
  debug& operator<<(const c&) {
    return *this;
  }
};
const long long int INF = 1e18;
void init() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
}
bool checkp2(long long int n) { return n && !(n & (n - 1)); }
long long int numsetbit(long long int n) {
  long long int count = 0;
  for (; n; count++) {
    n = n & (n - 1);
  }
  return count;
}
long long int powe(long long int x, long long int n, const long long int mod) {
  x = x % mod;
  long long int po = 1;
  while (n) {
    if (n & 1) {
      po = (po * x) % mod;
    }
    n = n >> 1;
    x = ((x % mod) * (x % mod)) % mod;
  }
  return po;
}
long long int powm(long long int x, long long int n) {
  long long int po = 1;
  while (n) {
    if (n & 1) {
      po = (po * x);
    }
    n = n >> 1;
    x = (x * x);
  }
  return po;
}
bool isPrime(long long int x) {
  if (x < 2) return 0;
  for (long long int y = 2; y * y <= x; y++) {
    if (x % y == 0) {
      return 0;
    }
  }
  return true;
}
const long long int mod = 1e9 + 7;
void solve() {
  long long int n;
  cin >> n;
  if (n == 2) {
    cout << "2" << endl;
    cout << "1 2" << endl;
    return;
  }
  cout << "2" << endl;
  cout << n << " " << n - 2 << endl;
  cout << n - 1 << " " << n - 1 << endl;
  for (long long int i = 0; i < n - 3; i++) {
    cout << n - i - 1 << " " << n - i - 3 << endl;
  }
}
int main() {
  init();
  long long int t = 1;
  cin >> t;
  for (long long int i = 1; i <= t; i++) {
    solve();
  }
}

```