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

public class test {

    public static void main(String[] args) {
        PrintWriter out = new PrintWriter(System.out);
        FastScanner fs=new FastScanner();
        int T=fs.nextInt();
        for (int tt=0; tt<T; tt++) {
            int n = fs.nextInt();
            int m = (int)Math.ceil(Math.sqrt(n));
            int ptr = m+1;
            
            int counter = n-2;

            int temp = n;
            while(temp > 2){
                temp = (int)Math.ceil(Math.sqrt(temp));
                counter++;
            }
            out.println(counter);

            while(n!=2){
                while(ptr < n){
                    out.println(ptr + " " + ++ptr);
                }
                out.println(n + " " + m);
                out.println(n + " " + m);
                n = m;
                m = (int)Math.ceil(Math.sqrt(n));
                ptr = m + 1;
            }
            // int ptr = n-1;

            // while(ptr >= 2){
            //     if((int)Math.ceil(Math.sqrt(n)) == ptr){
            //         out.println(n + " " + ptr);
            //         out.println(n + " " + ptr);
            //         n = ptr;

            //     }else{
            //         out.println(ptr + " " + (ptr+1));
            //     }
            //     ptr--;
            // }
            // out.println((n-3) + (int)Math.ceil(Math.log(n)/Math.log(2)));
            // while (ctr < n){
            //     out.println(ctr+ " " + ++ctr);
            // }
            // while(n > 1){
            //     out.println(temp + " " + 2);
            //     n = (int)Math.ceil((double)n/2);
            // }
        }
        out.close();
    }
    
    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[] readIntArray(int n) {
            int[] a=new int[n];
            for (int i=0; i<n; i++) a[i]=nextInt();
            return a;
        }
        long[] readLongArray(int n){
            long[] a=new long[n];
            for(int i=0; i<n; i++) a[i]=nextLong();
            return a;
        }
        long nextLong() {
            return Long.parseLong(next());
        }
    }
    
}
