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<iostream>
#include<bits/stdc++.h>
#define fast  ios_base::sync_with_stdio(false);cin.tie(NULL)
#define ll long long 
#define li long int 
#define pp pair<int,int> 
#define lb lower_bound
#define ub upper_bound
#define mk make_pair
#define pb push_back
#define llp(i,x,n) for(int i=x;i<n;i++)
#define fr first
#define ss second
#define mod 1000000007
using namespace std;
ll binpow(ll a, ll b) {
    a %= mod;  long long res = 1;
    while (b > 0) { if (b & 1)res = res * a % mod;
        a = a * a % mod; b >>= 1;}
       return res;}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int t;
    cin>>t;
    while(t--)
    {
        int n;
        cin>>n;
        int z=n;
        vector<pp>v;
        for(int i=n-1;i>=2;i--)
        {
            int x;
            if(z%i==0)
            x=z/i;
            else
            x=(z/i)+1;
            
            while(x>=i){
             v.pb(mk(n,i));
               z=x;
            if(z%i==0)
            x=z/i;
            else
            x=(z/i)+1;
            }
            v.pb(mk(i,n));
        }
        cout<<v.size()<<endl;
        llp(i,0,v.size())cout<<v[i].fr<<" "<<v[i].ss<<endl;
        
    }
      
}