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 arr(i,n) for(ll i=0;i<n;i++)
#define ARR(i,n) for(ll i=n-1;i>=0;i--)
#define f(i,a,b)  for(ll i=a;i<b;i++)
#define F(i,a,b)  for(ll i=b-1;i>=a;i--)
#define float double
#define PI 3.14159265358979323846
#define pb push_back
#define mp make_pair
#define vi vector<ll>
#define vc vector<char>
#define ceil(a,b) ceil(a*1.0/b)
#define vpii vector< pair<ll,ll> >
#define debug(x) cout<<(#x)<<" is "<<(x)<<endl
#define gcd(a,b) __gcd(a,b)
#define lcm(a,b) (a*(b/gcd(a,b)))
#define all(v) v.begin(),v.end()
#define inputarr(a,n) for(ll i=0;i<n;++i) cin>>a[i]
#define printarr(a,n) for(ll i=0;i<n;++i) cout<<a[i]<<" "
#define integer_to_string to_string 
#define IOS  ios::sync_with_stdio(0); cin.tie(0); cout.tie(0)
typedef long long ll;
using namespace std;
const ll  mod = 1000000000 + 7;
/*freopen("input.txt","r",stdin);
freopen("output.txt","w",stdout);*/
void solve()
{
    //code start her
    ll n;
    cin>>n;
    vpii v;
    f(i,3,n)
    {
        if(i!=10&&i!=4) v.pb(mp(i,i+1));
    }
    if(n>10)
    {
        ll temp=n;
        while(temp!=1)
        {
            v.pb(mp(n,10));
            temp=ceil(temp,10);
        }
        v.pb(mp(10,4));
        v.pb(mp(10,4));
        v.pb(mp(4,2));
        v.pb(mp(4,2));
    }
    else
    {
        ll temp=n;
        while(temp!=1)
        {
            v.pb(mp(n,2));
            temp=ceil(temp,2);
        }
        if(n>4) {v.pb(mp(4,2));
        v.pb(mp(4,2));}
    }
    cout<<v.size()<<endl;
    arr(i,v.size()) cout<<v[i].first<<" "<<v[i].second<<endl;
}
int main()
{
    int t;
    cin>>t;
    while(t--)
      solve();
    return 0;
}