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]. 
import static java.lang.System.exit; 

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.StringTokenizer;

public class Task4 {
	
	static void addRange(int start, int end, ArrayList<int[]> answ) {
		while (start < end) {
			int[] tmp = new int[2];
			tmp[0] = start;
			tmp[1] = start + 1;
			answ.add(tmp);
			start++;
		}
	}
	
	static void addToOne(int ind, int div, ArrayList<int[]> answ) {
		int val = ind;
		while(val > 1) {
			int[] tmp = new int[2];
			tmp[0] = ind;
			tmp[1] = div;
			answ.add(tmp);
			val = (val + div - 1) / div;
		}
	}
	
	static void printAnsw(ArrayList<int[]> answ) {
		out.println(answ.size());
		for (int i = 0; i < answ.size(); i++) {
			int[] tmp = answ.get(i);
			out.println(tmp[0] + " " + tmp[1]);
		}
	}
	
	static void solve() throws IOException{
		int n = scanInt();
		ArrayList<int[]> answ = new ArrayList<int[]>();
		if (n <= 8) {
			addRange(3, n, answ);
			addToOne(n, 2, answ);
			printAnsw(answ);
			
			return;
		}
		
		addRange(3, 8, answ);
		addRange(9, n, answ);
		addToOne(n, 8, answ);
		addToOne(8, 2, answ);
		printAnsw(answ);
	}
	
	
	
	static int scanInt() throws IOException {
		return Integer.parseInt(scanString());
	}
	
	static long scanLong() throws IOException {
		return Long.parseLong(scanString());
	}
	
	static String scanString() throws IOException {
		while (tok == null || !tok.hasMoreTokens()) {
			tok = new StringTokenizer(in.readLine());
		}
		
		return tok.nextToken();
	}
	
	static void printCase() {
		out.print("Case #" + test + ": ");
	}

	static void printlnCase() {
		out.println("Case #" + test + ":");
	}
	
	static BufferedReader in;
	static PrintWriter out;
	static StringTokenizer tok;
	static int test;
	
	public static void main(String[] args) throws IOException {
		try {
			in = new BufferedReader(new InputStreamReader(System.in));
			out = new PrintWriter(System.out);
			int tests = scanInt();
			for(test = 1; test <= tests; test++) {
				solve();
			}
			in.close();
			out.close();	
		} catch(Throwable e) {
			e.printStackTrace();
			exit(1);
		}
	}
}