#include <bits/stdc++.h>
using namespace std;
int dx[8] = {0, 0, 1, 1, -1, -1, 1, -1};
int dy[8] = {1, -1, 1, -1, 1, -1, 0, 0};
long long n, m, a[200005];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld", &n);
    printf("2\n");
    if (n == 2) {
      printf("1 2\n");
      continue;
    }
    if (n == 3) {
      printf("1 3\n");
      printf("2 2\n");
      continue;
    }
    m = 0;
    long long x = n;
    for (int i = 1; i <= n; i++) {
      a[i] = i;
    }
    while (m < n - 1) {
      long long y = x - 1;
      while ((a[x] + a[y]) % 2 != 0) {
        y--;
      }
      printf("%lld %lld\n", a[x], a[y]);
      a[y] = (a[x] + a[y]) / 2;
      x--;
      m++;
    }
  }
}

### Prompt

In Cpp, your task is to solve 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;
int dx[8] = {0, 0, 1, 1, -1, -1, 1, -1};
int dy[8] = {1, -1, 1, -1, 1, -1, 0, 0};
long long n, m, a[200005];
int main() {
  int t;
  scanf("%d", &t);
  while (t--) {
    scanf("%lld", &n);
    printf("2\n");
    if (n == 2) {
      printf("1 2\n");
      continue;
    }
    if (n == 3) {
      printf("1 3\n");
      printf("2 2\n");
      continue;
    }
    m = 0;
    long long x = n;
    for (int i = 1; i <= n; i++) {
      a[i] = i;
    }
    while (m < n - 1) {
      long long y = x - 1;
      while ((a[x] + a[y]) % 2 != 0) {
        y--;
      }
      printf("%lld %lld\n", a[x], a[y]);
      a[y] = (a[x] + a[y]) / 2;
      x--;
      m++;
    }
  }
}

```