#include <bits/stdc++.h>
using namespace std;
long long dx[] = {-1, 0, 1, 0};
long long dy[] = {0, 1, 0, -1};
long long spf[1];
void sieve() {
  long long i, j;
  for (i = 0; i < 1; i++) spf[i] = i;
  for (i = 4; i < 1; i += 2) spf[i] = 2;
  for (i = 3; i * i < 1; i++) {
    if (spf[i] == i) {
      for (j = i * i; j < 1; j += i) {
        if (spf[j] == j) spf[j] = i;
      }
    }
  }
}
bool isvowel(char c) {
  switch (c) {
    case 'a':
    case 'e':
    case 'i':
    case 'o':
    case 'u':
    case 'y':
      return true;
    default:
      return false;
  }
}
bool cmp(pair<long long, long long> x, pair<long long, long long> y) {
  return x.first > y.first;
}
bool isPrime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
long long sum(long long n) {
  long long ans = 0;
  while (n > 0) {
    long long d = n % 10;
    ans += d;
    n /= 10;
  }
  return ans;
}
int32_t 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;
    deque<long long> q;
    for (long long i = 1; i <= n; i++) {
      q.push_back(i);
    }
    cout << "2"
         << "\n";
    while (1) {
      if (q.size() == 1) {
        break;
      } else {
        long long a = q.back();
        q.pop_back();
        long long b = q.back();
        q.pop_back();
        cout << a << " " << b << "\n";
        long long x;
        if ((a + b) % 2) {
          x = ((a + b) / 2) + 1;
        } else {
          x = (a + b) / 2;
        }
        q.push_back(x);
      }
    }
  }
  return 0;
}

### 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;
long long dx[] = {-1, 0, 1, 0};
long long dy[] = {0, 1, 0, -1};
long long spf[1];
void sieve() {
  long long i, j;
  for (i = 0; i < 1; i++) spf[i] = i;
  for (i = 4; i < 1; i += 2) spf[i] = 2;
  for (i = 3; i * i < 1; i++) {
    if (spf[i] == i) {
      for (j = i * i; j < 1; j += i) {
        if (spf[j] == j) spf[j] = i;
      }
    }
  }
}
bool isvowel(char c) {
  switch (c) {
    case 'a':
    case 'e':
    case 'i':
    case 'o':
    case 'u':
    case 'y':
      return true;
    default:
      return false;
  }
}
bool cmp(pair<long long, long long> x, pair<long long, long long> y) {
  return x.first > y.first;
}
bool isPrime(long long n) {
  for (long long i = 2; i <= sqrt(n); i++) {
    if (n % i == 0) {
      return false;
    }
  }
  return true;
}
long long sum(long long n) {
  long long ans = 0;
  while (n > 0) {
    long long d = n % 10;
    ans += d;
    n /= 10;
  }
  return ans;
}
int32_t 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;
    deque<long long> q;
    for (long long i = 1; i <= n; i++) {
      q.push_back(i);
    }
    cout << "2"
         << "\n";
    while (1) {
      if (q.size() == 1) {
        break;
      } else {
        long long a = q.back();
        q.pop_back();
        long long b = q.back();
        q.pop_back();
        cout << a << " " << b << "\n";
        long long x;
        if ((a + b) % 2) {
          x = ((a + b) / 2) + 1;
        } else {
          x = (a + b) / 2;
        }
        q.push_back(x);
      }
    }
  }
  return 0;
}

```