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]. 
    #include<bits/stdc++.h>
    #define ii pair<int,int>
    #define s second
    #define f first
    #define ll long long
    #define ci pair<char,int>
    #define loob(i,n) for(int i=0 ;i<n ;++i)
    #define loob1(i,n) for(int i=1 ;i<=n ;++i)
    #define loobc(i,n) for(int i='a' ; i<'c' ;++i)
    #define rloob(i,n) for(int j=n-1 ;j>=0 ;--j)
    #define vi vector<int>
    #define sz size()
    #define pb push_back
    #define pp pop_back
    #define oo  0x3f3f3f3f3f3f3f3f
    #define intvector vector<int>
    #define llvector vector<long long>
    #define intset set<int>
    #define llset set<long long>
    #define intmap map<int,int>
    #define llmap map<long long ,long long>
    #define ilmap map<int ,long long>
    #define limap map<long long ,int >
using namespace std;
void GO(){
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
}
int dx[4]={1,-1,0,0};
int dy[4]={0,0,1,-1};
int main(){
    GO();
    int t; scanf("%d",&t);
    while(t--)
    {
        int n;scanf("%d",&n);
        vector<ii>v;
        if(n<=60)
        {
            for(int i=3 ;i<n ;++i)v.pb({i,i+1});
            int hold=n;
            while(hold!=1)
            {
                v.pb({n,2});
                hold=(hold+1)/2;
            }
            cout<<v.sz<<endl;
            for(auto x :v)cout<<x.f<<" "<<x.s<<endl;
            continue;
        }
        for(int i=3 ;i<n ;++i)
        {
            if(i==60||i==4)continue;
            v.pb({i,i+1});
        }
        int hold=n;
        while(hold!=1)
        {
            v.pb({n,60});
            hold=(hold+59)/60;
        }
        cout<<v.sz+5<<endl;
        for(auto x :v)cout<<x.f<<" "<<x.s<<endl;
        cout<<"60 4"<<endl;
        cout<<"60 4"<<endl;
        cout<<"60 4"<<endl;
        cout<<"4 2"<<endl;
        cout<<"4 2"<<endl;
    }
return 0;
}
