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]. 
/*******************************************************************************
* author  : SuryanshPrabhat
* created : 30/12/2020 01:26
* DO NOT GET STUCK ON ONE APPROACH!!! <Brute/Greedy/DP/PrefixSuffix/DFS/Insight>
*******************************************************************************/
import java.io.IOException;
import java.io.PrintWriter;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.io.BufferedOutputStream;
//import java.math.BigInteger;
import java.util.StringTokenizer;
import java.util.ArrayList;

public class d
{
    public static void main(String[] args)
    {
    	int T = in.nextInt();
    	while (T-- > 0)
    	{
            int n = in.nextInt();
            ArrayList<int[]> sequence = new ArrayList<>();
            int current = n-1, pivot = n;
            while (true)
            {
                int limit = (int)Math.sqrt(pivot);
                while (current > 2 && current > limit)
                {
                    sequence.add(new int[]{current, n});
                    current -= 1;
                }
                if (current == 2)
                {
                    break;
                }
                sequence.add(new int[]{n, current});
                pivot = (int)Math.ceil(1.0*pivot/current);
            }
            assert(current == 2);
            while (pivot != 1)
            {
                sequence.add(new int[]{n, current});
                pivot = (int)Math.ceil(1.0*pivot/current);
            }

            out.println(sequence.size());
            for (int[] op : sequence)
            {
                out.println(op[0] + " " + op[1]);
            }
    	}
        out.flush();
        out.close();
    }

    // Handle I/O
    static FastScanner in = new FastScanner();
    static PrintWriter out = new PrintWriter(new BufferedOutputStream(System.out));
    static class FastScanner
    {
        BufferedReader br;
        StringTokenizer st;
        public FastScanner()
        {
            br = new BufferedReader(new InputStreamReader(System.in));
            st = null;
        }

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

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