#include <bits/stdc++.h>
using namespace std;
const int mo = 1e9 + 7;
long long qp(long long a, long long b) {
  if (a == 0) return 0;
  long long res = 1;
  a %= mo;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mo;
    a = a * a % mo;
  }
  return res;
}
template <class T>
bool read(T& ret) {
  char c;
  int sgn;
  T bit = 0.1;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && c != '.' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  if (c == ' ' || c == '\n') {
    ret *= sgn;
    return 1;
  }
  while (c = getchar(), c >= '0' && c <= '9') ret += (c - '0') * bit, bit /= 10;
  ret *= sgn;
  return 1;
}
void solve() {
  priority_queue<int> q;
  vector<pair<int, int> > vp;
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) q.push(i);
  while (q.size() > 1) {
    int t = q.top();
    q.pop();
    int tt = q.top();
    q.pop();
    vp.push_back({t, tt});
    q.push((tt + t + 1) / 2);
  }
  cout << q.top() << endl;
  for (auto x : vp) {
    printf("%d %d\n", x.first, x.second);
  }
}
signed main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

### 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 mo = 1e9 + 7;
long long qp(long long a, long long b) {
  if (a == 0) return 0;
  long long res = 1;
  a %= mo;
  for (; b; b >>= 1) {
    if (b & 1) res = res * a % mo;
    a = a * a % mo;
  }
  return res;
}
template <class T>
bool read(T& ret) {
  char c;
  int sgn;
  T bit = 0.1;
  if (c = getchar(), c == EOF) return 0;
  while (c != '-' && c != '.' && (c < '0' || c > '9')) c = getchar();
  sgn = (c == '-') ? -1 : 1;
  ret = (c == '-') ? 0 : (c - '0');
  while (c = getchar(), c >= '0' && c <= '9') ret = ret * 10 + (c - '0');
  if (c == ' ' || c == '\n') {
    ret *= sgn;
    return 1;
  }
  while (c = getchar(), c >= '0' && c <= '9') ret += (c - '0') * bit, bit /= 10;
  ret *= sgn;
  return 1;
}
void solve() {
  priority_queue<int> q;
  vector<pair<int, int> > vp;
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) q.push(i);
  while (q.size() > 1) {
    int t = q.top();
    q.pop();
    int tt = q.top();
    q.pop();
    vp.push_back({t, tt});
    q.push((tt + t + 1) / 2);
  }
  cout << q.top() << endl;
  for (auto x : vp) {
    printf("%d %d\n", x.first, x.second);
  }
}
signed main() {
  int t;
  cin >> t;
  while (t--) solve();
  return 0;
}

```