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.*;
import java.util.*;

public class D {
    private static class InputReader {
        private final BufferedReader f;
        private StringTokenizer st;

        public InputReader(InputStream stream) {
            f = new BufferedReader(new InputStreamReader(stream));
            st = null;
        }

        public String next() {
            while (st == null || !st.hasMoreTokens()) {
                try {
                    st = new StringTokenizer(f.readLine());
                } catch (IOException e) {
                    throw new NoSuchElementException("No more elements to read");
                }
            }
            return st.nextToken();
        }

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

        public void close() throws IOException {
            f.close();
        }
    }

    private static Queue<String> ans;

    public static void main(String[] args) throws IOException {
        InputReader in = new InputReader(System.in);
        PrintWriter out = new PrintWriter(System.out);

        int t = in.nextInt();
        while (t-- > 0) {
            ans = new LinkedList<>();
            int n = in.nextInt();
            solve(n);
            out.println(ans.size());
            for (String pair : ans) {
                out.println(pair);
            }
        }

        in.close();
        out.close();
    }

    private static void solve(int y) {
        if (y == 2) return;
        int nextY = Math.max(1, (int) Math.sqrt(y) - 1);
        while (nextY < (y + nextY - 1) / nextY) {
            nextY++;
        }
        for (int i = nextY+1; i < y; i++) {
            ans.add(i+" "+y);
        }
        ans.add(y+" "+nextY);
        ans.add(y+" "+nextY);

        solve(nextY);
    }
}
