#include <bits/stdc++.h>
using namespace std;
const int M = 5e6 + 5;
const long long int mod = 1e9 + 7;
const long long int inf = 9e18;
const double pi = 3.14159265358979323846;
long long int t, n, m, x, sum, k, p, u = 0, t1 = 1;
string s, z, ans = "";
long long int P[10000005] = {0};
void sieve() {
  for (long long int i = 2; i < (10000005); i++)
    if (!P[i])
      for (long long int j = i * i; j < (10000005); j += i) P[j] = i;
}
long long int raised(long long int x, long long int v) {
  if (v == 0) return 1 % mod;
  if (v % 2 == 1) return (x * raised(x, v - 1)) % mod;
  long long int r = raised(x, v / 2);
  return (r * r) % mod;
}
bool prime(long long int n) {
  if (n == 1) return 1;
  for (int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int hcf(long long int a, long long int b) {
  if (a % b == 0) return b;
  return hcf(b, a % b);
}
bool comp(long long int a, long long int b) { return (a < b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> t1;
  while (t1--) {
    cin >> n;
    long long int a = n, b = n - 1, c;
    m = n;
    while (n > 1) {
      a = (a + b + 1) / 2;
      n--;
      b = n - 1;
    }
    cout << a << "\n";
    n = m;
    a = n;
    b = n - 1;
    while (n > 1) {
      cout << a << " " << b << "\n";
      a = (a + b + 1) / 2;
      n--;
      b = n - 1;
    }
  }
}

### 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;
const int M = 5e6 + 5;
const long long int mod = 1e9 + 7;
const long long int inf = 9e18;
const double pi = 3.14159265358979323846;
long long int t, n, m, x, sum, k, p, u = 0, t1 = 1;
string s, z, ans = "";
long long int P[10000005] = {0};
void sieve() {
  for (long long int i = 2; i < (10000005); i++)
    if (!P[i])
      for (long long int j = i * i; j < (10000005); j += i) P[j] = i;
}
long long int raised(long long int x, long long int v) {
  if (v == 0) return 1 % mod;
  if (v % 2 == 1) return (x * raised(x, v - 1)) % mod;
  long long int r = raised(x, v / 2);
  return (r * r) % mod;
}
bool prime(long long int n) {
  if (n == 1) return 1;
  for (int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int hcf(long long int a, long long int b) {
  if (a % b == 0) return b;
  return hcf(b, a % b);
}
bool comp(long long int a, long long int b) { return (a < b); }
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  ;
  cin >> t1;
  while (t1--) {
    cin >> n;
    long long int a = n, b = n - 1, c;
    m = n;
    while (n > 1) {
      a = (a + b + 1) / 2;
      n--;
      b = n - 1;
    }
    cout << a << "\n";
    n = m;
    a = n;
    b = n - 1;
    while (n > 1) {
      cout << a << " " << b << "\n";
      a = (a + b + 1) / 2;
      n--;
      b = n - 1;
    }
  }
}

```