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.util.*;
import java.io.*;

public class Main {
    static Scanner sc = new Scanner(System.in);
    static PrintWriter out = new PrintWriter(System.out);
    static BufferedReader reader = new BufferedReader(new InputStreamReader(System.in));
    //    long mod = 998244353;
    long mod = (long)1e9+7;
    int maxn = 200000;
    int[] dp;
    boolean flag = false;
    public static void main(String[] args) throws Exception {
        Main main = new Main();

        int t = sc.nextInt();
//        int t = 1;
        for(int i=1; i<=t; i++){
            main.solve();
        }
        out.flush();

    }

    void solve(){
        // n>=3
        if(!flag){
            flag = true;
            dp = new int[maxn+5];
            dp[1] = 1;
            for(int i=2; i<=maxn; i++){
                dp[i] = dp[i-1];
                while(!helper(i, dp[i])) dp[i]++;
            }
        }
        int n = neInt();
        int cur = n;
        boolean[] rec = new boolean[n+1];
        Deque<int[]> first = new LinkedList<>(), second = new LinkedList<>();
        while(cur>2){
            rec[cur] = true;
            int next = dp[cur];
            int[] temp = new int[]{cur, next};
            second.add(temp); second.add(temp);
            cur = next;
        }
        for(int i=3; i<=n; i++){
            if(rec[i]) continue;
            first.add(new int[]{i, i+1});
        }
        out.println(first.size()+second.size());
        
        for(int[] t:first){
            out.println(t[0]+" "+t[1]);
        }
        for(int[] t:second){
            out.println(t[0]+" "+t[1]);
        }

    }
    boolean helper(int n, int k){
        // return true if k >= ceiling(n/k);
        int res = n%k==0?n/k:n/k+1;
        return k>=res;
    }


    String neS(){
        return sc.next();
    }
    int neInt(){
        return Integer.parseInt(sc.next());
    }
    long neLong(){
        return Long.parseLong(sc.next());
    }

    int paIn(String s){return Integer.parseInt(s);}


}