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 JAVA):

Solution:
import java.io.*;
import java.util.*;

public class Main {
	static class FastReader
	{
	BufferedReader br;
	StringTokenizer st;

	public FastReader()
	{
	br = new BufferedReader(new
	InputStreamReader(System.in));
	}

	String next()
	{
	while (st == null || !st.hasMoreElements())
	{
	try
	{
	st = new StringTokenizer(br.readLine());
	}
	catch (IOException e)
	{
	e.printStackTrace();
	}
	}
	return st.nextToken();
	}

	int nextInt()
	{
	return Integer.parseInt(next());
	}

	long nextLong()
	{
	return Long.parseLong(next());
	}

	double nextDouble()
	{
	return Double.parseDouble(next());
	}

	String nextLine()
	{
	String str = "";
	try
	{
	str = br.readLine();
	}
	catch (IOException e)
	{
	e.printStackTrace();
	}
	return str;
	}
	}
	static int m = (int)(1e9+9);
	static long mod(long x) {
		return ((x%m + m)%m);
	}
	static long add(long a,long b) {
		return mod(mod(a)+mod(b));
	}
	static long mul(long a,long b) {
		return mod(mod(a) * mod(b));
	}
	
	public static void main(String[] args) {
		FastReader sc = new FastReader();
		int t = sc.nextInt();
		outer : while(t-->0) {
			int n = sc.nextInt();
			ArrayList<pair> a = new ArrayList<>();
			if(n<9) {
				for(int i=3;i<n;i++)
					a.add(new pair(i,n));
				int temp = n;
				while(n!=1) {
					a.add(new pair(temp,2));
					if(n%2==0) n/=2;
					else n=n/2+1;
				}
				System.out.println(a.size());
				for(pair p : a) System.out.println(p.a+" "+p.b);
				continue outer;
			}
			a.add(new pair(3,n));
			for(int i=5;i<n;i++) {
				if(i==8) continue;
				a.add(new pair(i,n));
			}
			for(int i=1;i<=6;i++) a.add(new pair(n,8));
			a.add(new pair(4,8));
			for(int i=1;i<=3;i++) a.add(new pair(8,2));
			System.out.println(a.size());
			for(pair p : a) System.out.println(p.a+" "+p.b);
		}
	}
	private static int sqrt(long x) {
		long l = 1,h = x;
		int ans = 0;
		while(l<=h) {
			long mid = (l+(h-l)/2);
			if(mid>x/mid) h = mid-1;
			else {
				ans = (int) mid;
				l = mid+1;
			}
		}
		return ans;
	}
}
class pair{
	int a;
	int b;
	pair(int a,int b){
		this.a = a;
		this.b = b;
	}
}