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.*;

public class CeilDivisions {
	public static void main(String[] args)  {
		FastScanner fs = new FastScanner();
		PrintWriter out = new PrintWriter(System.out);
		int testCase = fs.nextInt();
		for (int tc = 0; tc < testCase; tc++) {
			int n = fs.nextInt();
			List<Integer> ans = new ArrayList<>();
			for(int k = (int)Math.ceil(Math.sqrt(n)); n > 2; k = (int)Math.ceil(Math.sqrt(n))) {
				for(int i = k + 1; i < n; ++i) {
					ans.add(i);
					ans.add(n);
				}
				ans.add(n);
				ans.add(k);
				ans.add(n);
				ans.add(k);
				n = k;
			}
			out.println(ans.size() / 2);
			for(int i = 0; i < ans.size(); i += 2)
				out.println(ans.get(i) + " " + ans.get(i + 1));
		}
		out.close();
	}
	static void ruffleSort(int[] a) {
		int n = a.length;
		Random r = new Random();
		for(int i = 0; i < a.length; ++i) {
			int oi = r.nextInt(n), t = a[i];
			a[i] = a[oi];
			a[oi] = t;
		}
		Arrays.sort(a);
	}
	static void sort(int[] a)   {
		ArrayList<Integer> l = new ArrayList<>();
		for(int i : a) l.add(i);
		Collections.sort(l);
		for(int i = 0; i < a.length; ++i) a[i] = l.get(i);
	}
	static class FastScanner {
		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st = new StringTokenizer("");

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

		int nextInt() {
			return Integer.parseInt(next());
		}
		int[] readArray(int n) {
			int[] a = new int[n];
			for(int i = 0; i < n; ++i)
				a[i] = nextInt();
			return a;
		}
		long nextLong() {
			return Long.parseLong(next());
		}
	}
}