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 <iostream>
#include <cassert>
using namespace std;

#define NAX int(2e5+1)
int n;

void solve_less(){
	int num_steps = 0;
	
	for(int i=3;i<n;i++){
		num_steps++;
	}
	
	int cur = n;
	while(cur > 1){
		cur = (cur + 1) / 2;
		num_steps++;
	}
	
	assert(num_steps <= n+5);
	
	cout << num_steps << '\n';
	
	for(int i=3;i<n;i++){
		num_steps++;
		cout << i << ' ' << n << '\n';
	}
	
	cur = n;
	while(cur > 1){
		cur = (cur + 1) / 2;
		num_steps++;
		cout << n << ' ' << 2 << '\n';
	}
}

void solve(){
	cin >> n;
	if(n <= 12){
		solve_less();
		return;
	}
	
	int num_steps = 0;
	
	for(int i=3;i<n;i++){
		if(i != 12 and i != 4){
			num_steps++;
			// cout << i << ' ' << n << '\n';
		}
	}
	
	int cur = n;
	while(cur > 1){
		num_steps++;
		cur = (cur + 11) / 12;
		// cout << n << ' ' << 12 << '\n';
	}
	
	cur = 12;
	while(cur > 1){
		num_steps++;
		cur = (cur + 3) / 4;
		// cout << 12 << ' ' << 4 << '\n';
	}
	
	cur = 4;
	while(cur > 1){
		num_steps++;
		cur = (cur + 1) / 2;
		// cout << 4 << ' ' << 2 << '\n';
	}
	
	cout << num_steps << '\n';
	
	for(int i=3;i<n;i++){
		if(i != 12 and i != 4){
			num_steps++;
			cout << i << ' ' << n << '\n';
		}
	}
	
	cur = n;
	while(cur > 1){
		num_steps++;
		cur = (cur + 11) / 12;
		cout << n << ' ' << 12 << '\n';
	}
	
	cur = 12;
	while(cur > 1){
		num_steps++;
		cur = (cur + 3) / 4;
		cout << 12 << ' ' << 4 << '\n';
	}
	
	cur = 4;
	while(cur > 1){
		num_steps++;
		cur = (cur + 1) / 2;
		cout << 4 << ' ' << 2 << '\n';
	}
}

int main(){
	int test; cin >> test;
	while(test--){
		solve();
	}
}