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:
///Fix the cause, not the symptom.

#include<bits/stdc++.h>
using namespace std;
#define ll           long long
#define ull          unsigned long long
#define db           double
#define sf           scanf
#define pf           printf
#define pb           push_back
#define pi           acos(-1)
#define MOD          1000000007
#define pii          pair<ll,ll>
#define time_save    ios_base::sync_with_stdio(false); cin.tie(NULL);

#define D(x)         cerr << __LINE__ << ": " << #x << " = " << (x) << '\n'
#define DD(x,y)      cerr << __LINE__ << ": " << #x << " = " << x << "   " << #y << " = " << y << '\n'
#define DDD(x,y,z)   cerr << __LINE__ << ": " << #x << " = " << x << "   " << #y << " = " << y << "   " << #z << " = " << z << '\n'
#define DBG          cerr << __LINE__ << ": Hi" << '\n'



/*----------------------Graph Moves----------------*/
//const ll dx[]={+1,-1,+0,+0};
//const ll dy[]={+0,+0,+1,-1};
//const ll dx[]={+0,+0,+1,-1,-1,+1,-1,+1};   // Kings Move
//const ll dy[]={-1,+1,+0,+0,+1,+1,-1,-1};  // Kings Move
//const ll dx[]={-2, -2, -1, -1,  1,  1,  2,  2};  // Knights Move
//const ll dy[]={-1,  1, -2,  2, -2,  2, -1,  1}; // Knights Move
/*------------------------------------------------*/

/*-----------------------Bitmask------------------*/
//int Set(int N,int pos){return N=N  (1<<pos);}
//int reset(int N,int pos){return N= N & ~(1<<pos);}
//bool check(int N,int pos){return (bool)(N & (1<<pos));}
/*------------------------------------------------*/

int main()
{
    #ifndef ONLINE_JUDGE
       freopen("in.txt","r",stdin);
    ///   freopen("out.txt","w",stdout);
    #endif

    time_save;

    ll t;
    cin>>t;
    while(t--){
        ll n;
        cin>>n;

        vector<pii>ar;
        while(n>2){
            ll sq=sqrt(n);
            if(sq*sq!=n)
                sq++;

            for(ll i=sq+1;i<n;i++){
                ar.pb({i,i+1});
            }
            ar.pb({n,sq});
            ar.pb({n,sq});
            n=sq;
        }

        cout<<ar.size()<<endl;
        for(ll i=0;i<ar.size();i++)
            cout<<ar[i].first<<" "<<ar[i].second<<endl;
        //cout<<"-----------"<<endl;
    }
    return 0;
}