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.InputStream;
import java.io.InputStreamReader;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.List;
import java.util.StringTokenizer;

public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        Scanner in = new Main().new Scanner(inputStream);
        Main solver = new Main();
        solver.runn(in);
    }

    long MOD = 1000000007L;

    public void runn(Scanner in) {
        int t = in.ni();
        while (t-- > 0) {
            int n = in.ni();
            List<Pair> steps = new ArrayList<>();
            while (n != 2) {
                int s = (int) Math.ceil(Math.sqrt(n));
                for (int i = s + 1; i < n; i++)
                    steps.add(new Pair(i, i + 1));
                steps.add(new Pair(n, s));
                steps.add(new Pair(n, s));
                n = s;
            }
            System.out.println(steps.size());
            for (Pair p : steps)
                System.out.println(p.a + " " + p.b);
        }
    }

    class Scanner {
        BufferedReader br;
        StringTokenizer in;

        public Scanner(InputStream inputStream) {
            br = new BufferedReader(new InputStreamReader(inputStream));
        }

        public String next() {
            return hasMoreTokens() ? in.nextToken() : null;
        }

        public String rl() {
            try {
                return br.readLine();
            } catch (Exception e) {
                e.printStackTrace();
                return null;
            }
        }

        boolean hasMoreTokens() {
            while (in == null || !in.hasMoreTokens()) {
                String s = rl();
                if (s == null)
                    return false;
                in = new StringTokenizer(s);
            }
            return true;
        }

        public String nextString() {
            return next();
        }

        public int ni() {
            return Integer.parseInt(nextString());
        }

        public double nd() {
            return Double.parseDouble(nextString());
        }

        public long nl() {
            return Long.parseLong(nextString());
        }

        public int[] nia(int n) {
            int[] a = new int[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

        public long[] nla(int n) {
            long[] a = new long[n];
            for (int i = 0; i < n; i++) {
                a[i] = ni();
            }
            return a;
        }

    }

    int stoi(String a) {
        return Integer.parseInt(a);
    }

    void debug(int[] arr) {
        System.out.println(Arrays.toString(arr));
    }

    void debug(long[] arr) {
        System.out.println(Arrays.toString(arr));
    }

    void debug(int[][] arr) {
        for (int i = 0; i < arr.length; i++) {
            for (int j = 0; j < arr[i].length; j++)
                System.out.print(arr[i][j] + "\t");
            System.out.println();
        }
    }

    class Pair implements Comparable<Pair> {
        int a;
        int b;

        public Pair(int x, int y) {
            a = x;
            b = y;
        }

        public int compareTo(Pair other) {
            if (a == other.a)
                return Integer.compare(b, other.b);
            else
                return Integer.compare(a, other.a);
        }

        public int hashCode() {
            return a * 100007 + b;
        }

        public boolean equals(Object other) {
            Pair o = (Pair) other;
            return a == o.a && b == o.b;
        }
    }
}