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.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.PrintWriter;
import java.util.*;
import java.math.*;
public class code {

	public static void main(String[] args)throws IOException {
		FastReader sc = new FastReader();
		PrintWriter pw = new PrintWriter(System.out);
		int t = sc.nextInt();
		while (t-- > 0) {
			int n = sc.nextInt();
			ArrayList<Pair> list = new ArrayList<>();
			for (int i = 3; i < n; i++) {
				if (i == 4 || i == 16)
					continue;
				Pair p = new Pair();
				p.x = i;
				p.y = i + 1;
				list.add(p);
			}
			if (n > 16) {
				int k = n;
				while (k > 0) {
					Pair p = new Pair();
					p.x = n;
					p.y = 16;
					list.add(p);
					k = k / 16;
				}
			}
			if (n >= 16) {
				list.add(new Pair(16, 4));
				list.add(new Pair(16, 4));

			} else {
				int k = n;
				while (k > 1) {
					list.add(new Pair(n, 2));
					k = (k + 1) / 2;
				}
			}
			if (n >= 4) {
				list.add(new Pair(4, 2));
				list.add(new Pair(4, 2));
			}
			System.out.println(list.size());
			for (Pair p : list)
				System.out.println(p.x + " " + p.y);


		}





	}
}
class Pair {
	int x, y;
	public Pair() {}
	public Pair(int x, int y) {
		this.x = x;
		this.y = y;
	}

}






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;
	}
}