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]. 

#include<bits/stdc++.h>
using namespace std;

#define rep(i,a,b) for(i=a;i<b;i++)
#define repe(i,a,b) for(i=a;i>=b;i--)

#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0);

#define ll long long
#define ld long double
#define V vector<ll>
#define pp pair<ll,ll>
#define pb push_back
#define mp make_pair
#define Pr pair<ll,ll>
#define all(A) A.begin(),A.end()
#define allr(A) A.rbegin().A.rend()
#define ff first
#define ss second
#define len length()
int mod=1000000007;
ll g[200005];
ll i,n;
bool isPrime(ll n){
    if(n%2==0 || n%3==0) return false;
    if(n<=1)  return false;
    if(n<=3)  return true;
    for(ll i=5;i*i<=n;i=i+6)
        if(n%i==0||n%(i+2)==0)
           return false;
    return true;
}

void solve()
{   

    ll n,m,i;
    cin>>n;
    vector<pair<ll,ll>> v;
    ll a=n;
    if(n>8)
       {
          for(int i=3;i<n;i++)
          {
              if(i==8)
              continue;
              v.pb(mp(i,n));
          }
          int a=n;
          while(a>1)
          {
              v.pb(mp(n,8));
              if(a%8)
              {
                  a=a/8;
                  a++;
              }
              else
              {
                  a=a/8;
              }
          }
          v.pb(mp(8,2));
          v.pb(mp(8,2));
          v.pb(mp(8,2));
          //v.pb(mp(10,2));
          
       
       }
    else{
    for(ll i=3;i<n;i++)
    {
        v.pb(mp(i,n));
    }
    while(a>1)
    {
        v.pb(mp(n,2));
        ll x=a/2;
        if(a%2)
        x++;
        a=x;
    }
    }
    cout<<v.size()<<"\n";
    for(auto x:v)
    {
        cout<<x.first<<" "<<x.second;
        cout<<"\n";
    }
    


}







int main()
{
    fast;
    ll test;
    cin>>test;
    ll i;
    while(test--){
    solve();
   // cout<<"\n";
    }
}

