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>
#define fast ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define ll long long int
#define pb push_back
#define mp make_pair
#define sp " "
#define mod (1000*1000*1000+7)
#define ff first
#define ss second
#define sz(x) x.size()
#define all(x) x.begin(),x.end()
#define pl pair<ll,ll>
#define vi vector<ll>
#define vii vector<pl>
#define ii pair<int,int>
#define val(x) cout<<#x<<" is "<<x<<endl
bool flag=true;
using namespace std;
const double epsilon = 0.0000001f;
const double PI = 3.141592653589793238460;
const ll cons=1e5+1;
const ll inf = 998244353;
//cout<<"Case #"<<te++<<": "<<

int main()
{
	fast;
	ll T=1,te=1;
	cin>>T;
	while(T--)
	{
	    ll i=0,j=0,ans=0,x=0,y=0,z=0;
	    ll n,k=0,m=0,mn=INT_MAX,mx=INT_MIN;
	    cin>>n;
	    vii p;
	    while(n>2)
        {
            int x=ceil(sqrt(n));
            for(i=x+1;i<n;i++)
            {
                p.pb(mp(i,n));
            }
            p.pb(mp(n,x));
            p.pb(mp(n,x));
            n=x;
        }
        cout<<sz(p)<<endl;
        for(auto it:p)
        {
            cout<<it.ff<<sp<<it.ss<<endl;
        }
        //val(ans);
	}
	return 0;
}
/*test cases
20 5
0 1 9 10 2 2 7 2 10 7 2 0 8 0 9 8 1 0 4 4
*/