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;
#define pii pair<int,int>
#define int long long

const int MOD = 1e9+7;
const int MAX = 2e8+1;
int t,n,m,k;
int rp[200001];
vector<pii > ans;
vector<int> cs;

signed main()
{
    ios::sync_with_stdio(false);
    cout.tie(nullptr);
    cin.tie(nullptr);
    cin>>t;
    for(int _ = 0 ; _ < t ; _++)
    {
        ans.clear();
        cs.clear();
        cin>>n;
        for(int i = 0 ; i < n ; i++)
        {
            rp[i]=0;
        }
        rp[1]=1;
        int cc = 2;
        while(cc<n)
        {
            rp[cc]=1;
            cs.push_back(cc);
            cc=cc*cc;
        }
        cs.push_back(n);
        for(int i = 1 ; i < n ; i++)
        {
            if(rp[i]==0)
            {
                ans.push_back({i,n});
            }
        }
        for(unsigned int i = cs.size()-1 ; i>0 ; i--)
        {
            ans.push_back({cs[i],cs[i-1]});
            ans.push_back({cs[i],cs[i-1]});
        }
        cout<<ans.size()<<"\n";
        for(unsigned int i = 0 ; i < ans.size() ; i++)
        {
            cout<<ans[i].first<<" "<<ans[i].second<<"\n";
        }
        
    }
    
}