#include<bits/stdc++.h>
using namespace std;
 
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define endl "\n"
#define int long long
#define vi vector<int> 
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define B begin()
#define E end()
 
 
int32_t main()
{
IOS;
int t;
cin >> t;
while(t--)
{
    int n;
    cin >> n;
    vector<pair<int,int>> v;
    if(n<=8)
    {
        for(int i=3;i<n;i++)
        {
            v.pb({i,n});
        }
        int k=n;
        while(n!=1)
        {
            v.pb({k,2});
            n = (n+1)/2;
        }
    }
    else
    {
        for(int i=3;i<n;i++)
        {
            if(i==8) continue;
            v.pb({i,n});
        }
        int k=n;
        while(n!=1)
        {
            v.pb({k,8});
            n = (n+7)/8;
        }
        v.pb({8,2});
        v.pb({8,2});
        v.pb({8,2});

    }
    cout << v.size() << endl;
    for(auto i: v)
    {
        cout << i.ff << " " << i.ss << endl;
    }
}


return 0;
}
### Prompt

Create a solution in Cpp for the following 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]. 

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
 
#define IOS ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define endl "\n"
#define int long long
#define vi vector<int> 
#define pb push_back
#define mp make_pair
#define ff first
#define ss second
#define B begin()
#define E end()
 
 
int32_t main()
{
IOS;
int t;
cin >> t;
while(t--)
{
    int n;
    cin >> n;
    vector<pair<int,int>> v;
    if(n<=8)
    {
        for(int i=3;i<n;i++)
        {
            v.pb({i,n});
        }
        int k=n;
        while(n!=1)
        {
            v.pb({k,2});
            n = (n+1)/2;
        }
    }
    else
    {
        for(int i=3;i<n;i++)
        {
            if(i==8) continue;
            v.pb({i,n});
        }
        int k=n;
        while(n!=1)
        {
            v.pb({k,8});
            n = (n+7)/8;
        }
        v.pb({8,2});
        v.pb({8,2});
        v.pb({8,2});

    }
    cout << v.size() << endl;
    for(auto i: v)
    {
        cout << i.ff << " " << i.ss << endl;
    }
}


return 0;
}
```