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>
#define N 200005
using namespace std;
int T;
int a[N],n,now,sum,x[N],y[N],nn[N];
bool tag[N];
void work(){
	memset(tag,0,sizeof(tag));
	scanf("%d",&n);
	if(n<=2){
		printf("0\n"); return ;
	}
	now=sum=0;
	for(int i=n;i>=2;i--){
		if(!now) { nn[++sum]=i; now=i; tag[i]=1; continue; }
		if(1ll*(i-1)*(i-1)<now){
			nn[++sum]=i; now=i; tag[i]=1;
			continue;
		}
	}
	if(nn[sum]!=2) tag[2]=1,nn[++sum]=2;
	printf("%d\n",n+sum-1-2);
	for(int i=n;i>=2;i--) if(!tag[i]) printf("%d %d\n",i,n);
	for(int i=1;i<sum;i++){
		printf("%d %d\n",nn[i],nn[i+1]);
		printf("%d %d\n",nn[i],nn[i+1]);
	}
}
int main(){
	cin>>T;
	while(T--){
		work();
	}
}