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;
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n;
        cin >> n;
        int n1 = n;
       int f[200000 +2]={0,};
        vector<int> temp;
        while (n1 !=2)
        {
            n1 = (int)sqrt(n1) + 1;
           f[n1] = 1;
            temp.push_back(n1); 
         // cout<<n1<<endl;
        }
        vector<pair<int, int>> ans;
        for (int i = n - 1; i > 2; i--)
        {
           if (f[i] == 0)
                ans.push_back(make_pair(i, n));
        }
        ans.push_back(make_pair(n, temp[0]));
        ans.push_back(make_pair(n, temp[0]));

        int p = temp[0];
        for (int i = 1; i < temp.size(); i++)
        {
            ans.push_back(make_pair(p, temp[i]));
            ans.push_back(make_pair(p, temp[i]));
            p=temp[i];
        }
        // if(temp.size() > 1)
        // ans.push_back(make_pair(p, 2));
        cout<<ans.size()<<endl;
        for(int i=0;i<ans.size();i++)
        cout<<ans[i].first<<" "<<ans[i].second<<endl;
    }
}