Problem: You have an array a_1, a_2, ..., a_n where a_i = i.

In one step, you can choose two indices x and y (x ≠ y) and set a_x = \left⌈ (a_x)/(a_y) \right⌉ (ceiling function).

Your goal is to make array a consist of n - 1 ones and 1 two in no more than n + 5 steps. Note that you don't have to minimize the number of steps.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first and only line of each test case contains the single integer n (3 ≤ n ≤ 2 ⋅ 10^5) — the length of array a.

It's guaranteed that the sum of n over test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case, print the sequence of operations that will make a as n - 1 ones and 1 two in the following format: firstly, print one integer m (m ≤ n + 5) — the number of operations; next print m pairs of integers x and y (1 ≤ x, y ≤ n; x ≠ y) (x may be greater or less than y) — the indices of the corresponding operation.

It can be proven that for the given constraints it's always possible to find a correct sequence of operations.

Example

Input


2
3
4


Output


2
3 2
3 2
3
3 4
4 2
4 2

Note

In the first test case, you have array a = [1, 2, 3]. For example, you can do the following: 

  1. choose 3, 2: a_3 = \left⌈ (a_3)/(a_2) \right⌉ = 2 and array a = [1, 2, 2]; 
  2. choose 3, 2: a_3 = \left⌈ 2/2 \right⌉ = 1 and array a = [1, 2, 1]. 

You've got array with 2 ones and 1 two in 2 steps.

In the second test case, a = [1, 2, 3, 4]. For example, you can do the following: 

  1. choose 3, 4: a_3 = \left⌈ 3/4 \right⌉ = 1 and array a = [1, 2, 1, 4]; 
  2. choose 4, 2: a_4 = \left⌈ 4/2 \right⌉ = 2 and array a = [1, 2, 1, 2]; 
  3. choose 4, 2: a_4 = \left⌈ 2/2 \right⌉ = 1 and array a = [1, 2, 1, 1]. 

Solution (in CPP):

Solution:
#include<bits/stdc++.h>
//#include<bits/extc++.h>
#define ll long long
//#define int long long
//#define int _int128
//#define int _uint128_t
#define ull unsigned long long
#define MMST(x,y) memset(x,y,sizeof(x))

using namespace std;
//using namespace _gnu_pbds;

const int INF=0x3f3f3f3f;

int read()
{
	char c;
	int num,f=1;
	while(c=(char)getchar(),!isdigit(c))if(c=='-')f=-1;
	num=(int)(c-'0');
	while(c=(char)getchar(),isdigit(c))num=num*10+(int)(c-'0');
	return num*f;
}
void prt(int x)
{
 	if(x<0)
 	{
 		putchar('-');
 		x=-x;
	 }
	 if(x>9)prt(x/10);
	 putchar(x%10+'0');
}

int n;
int aa[200010],bb[200010],ans;
void work()
{
	n=read();
	ans=0;
	while(n>2)
	{
		int m=ceil(sqrt(n));
		for(int i=m+1;i<n;i++)
		{
			ans++;
			aa[ans]=i;
			bb[ans]=i+1;
		}
		ans++;
		aa[ans]=n;
		bb[ans]=m;
		ans++;
		aa[ans]=n;
		bb[ans]=m;
		n=m;
	}
	printf("%d\n",ans);
	for(int i=1;i<=ans;i++)
	{
		printf("%d %d\n",aa[i],bb[i]);
	}
	return;
}

signed main()
{
	int T=read();
	for(int Case=1;Case<=T;Case++)
	{
		//printf("Case #%d: ",Case);
		work();
	}
	return 0;
}