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 unsigned long long ll;

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    int tc;
    cin >> tc;
    while(tc--){
        int n;
        cin >> n;
        set<int> spec = {2,4,16,256,65536};
        vector<pair<int,int>> res;
        vector<int> arr(n+1);
        for(int i = 0;i <= n;i++)arr[i] = i;
        for(int i = n-1;i > 1;i--){
            if(spec.find(i) == spec.end()){
                res.push_back({n,i});
                arr[i] = arr[i]/n + arr[i]%n?1:0;
            }
        }
        int end = n;
        

        while(end <= *spec.rbegin())spec.erase(*spec.rbegin());
        int k = *spec.rbegin();
        res.push_back({k,end});
        int End = end;
        end = end/k + ((end%k)?1:0);
        arr[End] = arr[End]/(*spec.rbegin()) + arr[End]%(*spec.rbegin())?1:0;
        if(end != 1){
            res.push_back({*spec.rbegin(),End});
            arr[End] = arr[End]/(*spec.rbegin()) + arr[End]%(*spec.rbegin())?1:0;
            end = end/(*spec.rbegin()) + end%(*spec.rbegin())?1:0;
        }
        assert(end == 1);
        while(spec.size() > 1){
            end = *spec.rbegin();
            spec.erase(*spec.rbegin());
            arr[end] = arr[end]/(*spec.rbegin()) + arr[end]%(*spec.rbegin())?1:0;
            arr[end] = arr[end]/(*spec.rbegin()) + arr[end]%(*spec.rbegin())?1:0;
            res.push_back({*spec.rbegin(),end});
            res.push_back({*spec.rbegin(),end});
        }
        cout << res.size() << "\n";
        for(auto& x:res)cout << x.second << " " << x.first << "\n";
    }           
}