#include <bits/stdc++.h>
using namespace std;
const double pi = acos(-1);
bitset<200000> vis(0);
vector<bool> prime;
void primes(long long n) {
  prime.resize(n + 1);
  prime.assign(n + 1, true);
  prime[0] = 0;
  prime[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (prime[i]) {
      for (long long j = (i * i); j < n; j = j + i) prime[j] = false;
    }
  }
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long powM(long long x, long long y, long long m) {
  long long ans = 1, r = 1;
  x = x % m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans = ans * x;
      ans = ans % m;
    }
    r = r << 1;
    x = x * x;
    x = x % m;
  }
  return ans;
}
long long powN(long long a, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = ans * a;
    a = a * a;
    n >>= 1;
  }
  return ans;
}
long long modinv(long long a, long long m) { return powM(a, m - 2, m); }
long long ncr(long long n, long long r, long long m) {
  if (n < r) return 0;
  if (n == r || r == 0) return 1;
  if (r == 1) return n;
  if (n - r < r) return ncr(n, n - r, m);
  long long a = n;
  long long b = r;
  for (long long i = 1; i < r; i++) {
    a = (a * (n - i)) % m;
    b = (b * (r - i)) % m;
  }
  b = modinv(b, m);
  a = (a * b) % m;
  return a;
}
long long countbits(long long n) {
  long long cot = 0;
  while (n) {
    cot++;
    n >>= 1;
  }
  return cot;
}
long long check_prime(long long n) {
  long long ans = 1;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans = 0;
      break;
    }
  }
  if (n == 0) ans = 0;
  return ans;
}
int main() {
  fast();
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cout << 2 << "\n";
    vector<long long> a;
    for (long long i = 0; i < n; i++) a.push_back(i + 1);
    long long tt = n - 1;
    while (tt--) {
      long long k = a.size();
      if (k == 2) {
        cout << a[0] << " " << a[1] << "\n";
        continue;
      }
      if (a[k - 1] % 2 == a[k - 2] % 2) {
        long long m = (a[k - 1] + a[k - 2]) / 2;
        cout << a[k - 1] << " " << a[k - 2] << "\n";
        a.pop_back();
        a.pop_back();
        a.push_back(m);
      } else {
        long long m = (a[k - 1] + a[k - 3]) / 2;
        long long j = a[k - 2];
        cout << a[k - 1] << " " << a[k - 3] << "\n";
        a.pop_back();
        a.pop_back();
        a.pop_back();
        a.push_back(j);
        a.push_back(m);
      }
    }
  }
  return 0;
}

### Prompt

Your task is to create 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;
const double pi = acos(-1);
bitset<200000> vis(0);
vector<bool> prime;
void primes(long long n) {
  prime.resize(n + 1);
  prime.assign(n + 1, true);
  prime[0] = 0;
  prime[1] = 0;
  for (long long i = 2; i * i <= n; i++) {
    if (prime[i]) {
      for (long long j = (i * i); j < n; j = j + i) prime[j] = false;
    }
  }
}
void fast() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
}
long long gcd(long long x, long long y) {
  if (x == 0) return y;
  return gcd(y % x, x);
}
long long powM(long long x, long long y, long long m) {
  long long ans = 1, r = 1;
  x = x % m;
  while (r > 0 && r <= y) {
    if (r & y) {
      ans = ans * x;
      ans = ans % m;
    }
    r = r << 1;
    x = x * x;
    x = x % m;
  }
  return ans;
}
long long powN(long long a, long long n) {
  long long ans = 1;
  while (n > 0) {
    if (n & 1) ans = ans * a;
    a = a * a;
    n >>= 1;
  }
  return ans;
}
long long modinv(long long a, long long m) { return powM(a, m - 2, m); }
long long ncr(long long n, long long r, long long m) {
  if (n < r) return 0;
  if (n == r || r == 0) return 1;
  if (r == 1) return n;
  if (n - r < r) return ncr(n, n - r, m);
  long long a = n;
  long long b = r;
  for (long long i = 1; i < r; i++) {
    a = (a * (n - i)) % m;
    b = (b * (r - i)) % m;
  }
  b = modinv(b, m);
  a = (a * b) % m;
  return a;
}
long long countbits(long long n) {
  long long cot = 0;
  while (n) {
    cot++;
    n >>= 1;
  }
  return cot;
}
long long check_prime(long long n) {
  long long ans = 1;
  for (long long i = 2; i * i <= n; i++) {
    if (n % i == 0) {
      ans = 0;
      break;
    }
  }
  if (n == 0) ans = 0;
  return ans;
}
int main() {
  fast();
  long long t = 1;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    cout << 2 << "\n";
    vector<long long> a;
    for (long long i = 0; i < n; i++) a.push_back(i + 1);
    long long tt = n - 1;
    while (tt--) {
      long long k = a.size();
      if (k == 2) {
        cout << a[0] << " " << a[1] << "\n";
        continue;
      }
      if (a[k - 1] % 2 == a[k - 2] % 2) {
        long long m = (a[k - 1] + a[k - 2]) / 2;
        cout << a[k - 1] << " " << a[k - 2] << "\n";
        a.pop_back();
        a.pop_back();
        a.push_back(m);
      } else {
        long long m = (a[k - 1] + a[k - 3]) / 2;
        long long j = a[k - 2];
        cout << a[k - 1] << " " << a[k - 3] << "\n";
        a.pop_back();
        a.pop_back();
        a.pop_back();
        a.push_back(j);
        a.push_back(m);
      }
    }
  }
  return 0;
}

```