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:
#include<bits/stdc++.h>
using namespace std;
 
typedef vector <int> vi;
typedef pair< int ,int > pii;
#define endl "\n"
#define sd(val) scanf("%d",&val)
#define ss(val) scanf("%s",&val)
#define sl(val) scanf("%lld",&val)
#define debug(val) printf("check%d\n",val)
#define all(v) v.begin(),v.end()
#define PB push_back
#define MP make_pair
#define FF first
#define SS second
#define ll long long
#define MOD 1000000007
#define clr(val) memset(val,0,sizeof(val))
#define what_is(x) cerr << #x << " is " << x << endl;
#define OJ \
    freopen("input.txt", "r", stdin); \
    freopen("output.txt", "w", stdout);
#define FIO ios_base::sync_with_stdio(false);cin.tie(NULL);cout.tie(NULL);
 
 
int main()
{
  //FIO;
 int t; cin>>t;
 while(t--)
 {
     int n;
     cin>>n;
     vector<pair<int,int>>ans;
     if(n>16)
     {
         for(int i=17;i<n;i++)
         {
             ans.push_back({i,n});
         }
         int k=n;
         while (k>1)
         {
             ans.push_back({n,16});
             k =  ceil(k/16.0);
         }
         for(int i=9;i<16;i++)
        {
            ans.push_back({i,16});
        }
        ans.push_back({16,8});ans.push_back({16,8});
        for(int i=5;i<8;i++)
        {
            ans.push_back({i,8});
        }
        ans.push_back({8,4});ans.push_back({8,4});
        ans.push_back({3,4});
        ans.push_back({4,2});ans.push_back({4,2});
        
     
     }else
     {   

        for(int i=3;i<n;i++)
        {
            ans.push_back({i,n});
        }
        int k=n;
        while(k>1)
        {
            ans.push_back({n,2});
            k= ceil(k/2.0);
        }
     }
     cout<<ans.size()<<endl;
        for(int i=0;i< ans.size();i++)
        {
            cout<<ans[i].FF<<" "<<ans[i].SS<<endl;
        }
 }
 
 
return 0;
   
}