#include <bits/stdc++.h>
using namespace std;
priority_queue<long long int, vector<long long int>, greater<long long int> >
    ti;
vector<long long int> p[200005], f(300005, 0), b(200005, 1), r(200005, 0);
map<pair<long long int, long long int>, long long int> mp;
set<pair<long long int, long long int> > s;
set<long long int> st;
map<long long int, long long int> np, vt;
queue<long long int> qy;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 998244353;
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res % 998244353;
}
void fact(long long int i) {
  f[0] = 1;
  for (long long int k = 1; k <= i; k++) {
    (f[k] = f[k - 1] * k) %= 998244353;
  }
}
long long int isprime(long long int n) {
  if (n == 1) return 0;
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int find(long long int x) {
  if (f[x] == x)
    return x;
  else
    return f[x] = find(f[x]);
}
bool cmp(long long int x, long long int y) { return x < y; }
long long int comb(long long int i, long long int j) {
  long long int k = f[i];
  long long int g = (f[j] * (f[i - j])) % 998244353;
  long long int h = bpow(g, 998244353 - 2);
  return (k * h) % 998244353;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    cout << 2 << "\n";
    if (n == 2) {
      cout << 1 << " " << 2 << "\n";
    }
    long long int c = 0;
    while (n - 2 >= 1) {
      cout << n << " " << n - 2 << "\n";
      c++;
      n--;
      if (c == 1) {
        cout << n << " " << n << "\n";
      }
    }
  }
}

### 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;
priority_queue<long long int, vector<long long int>, greater<long long int> >
    ti;
vector<long long int> p[200005], f(300005, 0), b(200005, 1), r(200005, 0);
map<pair<long long int, long long int>, long long int> mp;
set<pair<long long int, long long int> > s;
set<long long int> st;
map<long long int, long long int> np, vt;
queue<long long int> qy;
long long int gcd(long long int a, long long int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
long long int bpow(long long int a, long long int b) {
  long long int res = 1;
  while (b > 0) {
    if (b & 1) res = (res * a) % 998244353;
    a = (a * a) % 998244353;
    b >>= 1;
  }
  return res % 998244353;
}
void fact(long long int i) {
  f[0] = 1;
  for (long long int k = 1; k <= i; k++) {
    (f[k] = f[k - 1] * k) %= 998244353;
  }
}
long long int isprime(long long int n) {
  if (n == 1) return 0;
  for (long long int i = 2; i <= sqrt(n); i++)
    if (n % i == 0) return 0;
  return 1;
}
long long int find(long long int x) {
  if (f[x] == x)
    return x;
  else
    return f[x] = find(f[x]);
}
bool cmp(long long int x, long long int y) { return x < y; }
long long int comb(long long int i, long long int j) {
  long long int k = f[i];
  long long int g = (f[j] * (f[i - j])) % 998244353;
  long long int h = bpow(g, 998244353 - 2);
  return (k * h) % 998244353;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long int t;
  cin >> t;
  while (t--) {
    long long int n;
    cin >> n;
    cout << 2 << "\n";
    if (n == 2) {
      cout << 1 << " " << 2 << "\n";
    }
    long long int c = 0;
    while (n - 2 >= 1) {
      cout << n << " " << n - 2 << "\n";
      c++;
      n--;
      if (c == 1) {
        cout << n << " " << n << "\n";
      }
    }
  }
}

```