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.util.ArrayList;
import java.util.Scanner;


public class A {
    private static Scanner scanner=new Scanner(System.in);
    static int t= scanner.nextInt();
    static ArrayList<Pair<Integer,Integer>> a=new ArrayList<>();
    public static void main(String[] args) {
        scanner.nextLine();

        for(int tt=0;tt<t;tt++) {
            int n= scanner.nextInt();
            solve(n);
            System.out.println(a.size());
            for(int i=0;i<a.size();i++)
            {
                System.out.println(a.get(i).first+" "+a.get(i).second);
            }
            a.clear();

        }
    }
    public static void solve(int n)
    {
        if(n==2)return;

        int m= (int) Math.sqrt(n);
        while(m*m-m+1<n)m++;
        for(int i=m+1;i<n;i++)
            a.add(new Pair<>(i,n));
        a.add(new Pair<>(n,m));
        a.add(new Pair<>(n,m));
        solve(m);

    }


    public static long __gcd(long a,long b){
        if(b>a)
        {
            long temp=a;a=b;b=temp;
        }
        if(b==0)return a;
        return __gcd(b,a%b);
    }

    public static void print(int a[],int n) {
        for (int i = 0; i < n; i++) {
            System.out.print(a[i] + " ");
        }
        System.out.println();
    }

    static int LowerBound(int a[], int x) {
        int l=-1,r=a.length;
        while(l+1<r) {
            int m=(l+r)>>>1;
            if(a[m]>=x) r=m;
            else l=m;
        }
        return r;
    }
    static int UpperBound(int a[], int x) {// x is the key or target value
        int l=-1,r=a.length;
        while(l+1<r) {
            int m=(l+r)>>>1;
            if(a[m]<=x) l=m;
            else r=m;
        }
        return l+1;
    }

}
class DSU {
    long n;
    long[] rank,parent,comp,iscircle;
    public DSU(long n) {
        this.n = n;
        rank= new long[(int) n];
        parent= new long[(int) n];
        comp= new long[(int) n];
        iscircle=new long[(int) n];

        for(int i=0;i<n;i++)
        {
            rank[i]=0;iscircle[i]=0;
            parent[i]=i;comp[i]=1;
        }
    }

    public long find(long n)
    {
        long x=n;
        while(x!=parent[(int) x])
            x=parent[(int) x];
        return x;
    }

    public void uni(long x, long y)
    {
        long px=find(x);long py=find(y);
        if(px==py)return;
        if(rank[(int) px]>rank[(int) py])
        {
            parent[(int) py]=px;comp[(int) px]+=comp[(int) py];
        }
        else if(rank[(int) px]<rank[(int) py])
        {
            parent[(int) px]=py;comp[(int) py]+=comp[(int) px];
        }
        else
        {
            parent[(int) py]=px;
            rank[(int) px]++;
            comp[(int) px]+=comp[(int) py];
        }
    }
}



class Pair <T extends Comparable,T1 extends Comparable> implements Comparable<Pair>{
    T first;
    T1 second;

    public Pair(T first, T1 second) {
        this.first = first;
        this.second = second;
    }

    @Override
    public int compareTo(Pair o) {
        if(this.first.compareTo(o.first)!=0)
        {
            return this.first.compareTo(o.first);
        }
        else
        {
            return this.second.compareTo(o.second);
        }
    }

    @Override
    public String toString() {
        return "Pair{" +
                "first=" + first +
                ", second=" + second +
                '}';
    }
}