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 CPP):

Solution:
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
//_CRT_SECURE_NO_WARNINGS
using namespace std;
#define IOS ios::sync_with_stdio(0);
#define int long long 
#define pb push_back
#define f first
#define si second
#define mp make_pair
signed main(){
  int q=1;
   cin>>q;
   while(q--)
   {
       int n;
    cin>>n;
    vector<pair<int,int>>p;
    for(int i=3;i<n;i++)
    {
        if(i!=4&&i!=8)
        {
            p.pb(mp(i,n));
        }
    }
    if(n>=8)
    {
        if(n!=8)
        {
        int temp=n;
        while(temp>1)
        {
            p.pb(mp(n,8));
            temp=temp/8;
            if(temp%8!=0)
            temp++;
        }
        }
    
    p.pb(mp(8,4));
    p.pb(mp(8,4));
    p.pb(mp(4,2));
    p.pb(mp(4,2));
    }
    else if(n>=4)
    {
        if(n!=4)
        {
        p.pb(mp(n,4));
        p.pb(mp(n,4));
        }
    p.pb(mp(4,2));
    p.pb(mp(4,2));
    }
    else
    {
        p.pb(mp(3,2));
        p.pb(mp(3,2));
    }
    cout<<p.size()<<endl;
    for(int i=0;i<p.size();i++)
    {
        cout<<p[i].f<<" "<<p[i].si<<endl;
    }
    
   }
}