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;
#define ll long long
#define dd double
#define endl "\n"
#define pb push_back
#define all(v) v.begin(),v.end()
#define mp make_pair
#define fi first
#define se second
#define vll vector<ll>
#define pll pair<ll,ll>
#define fo(i,n) for(int i=0;i<n;i++)
#define fo1(i,n) for(int i=1;i<=n;i++)
ll mod=1000000007;
ll n,k,t,m,q,flag=0;
ll power(ll a,ll b) {ll res=1;a%=mod; assert(b>=0); for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
// #include <ext/pb_ds/assoc_container.hpp> 
// #include <ext/pb_ds/tree_policy.hpp> 
// using namespace __gnu_pbds; 
// #define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update>
// ordered_set s ; s.order_of_key(a) -- no. of elements strictly less than a
// s.find_by_order(i) -- itertor to ith element (0 indexed)
ll min(ll a,ll b){if(a>b)return b;else return a;}
ll max(ll a,ll b){if(a>b)return a;else return b;}
ll gcd(ll a , ll b){ if(b > a) return gcd(b , a) ; if(b == 0) return a ; return gcd(b , a%b) ;}
int main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    #ifdef NOOBxCODER
    freopen("input.txt", "r", stdin);
    freopen("output.txt", "w", stdout);
    #else 
    #define NOOBxCODER 0
    #endif
    cin>>t;
    //t=1;
    while(t--){
        ll count=0;
        cin>>n;
        if(n<20){
        m=n;
        while(n>1){
            n= (n+1)/2;
            count++;
        }
        n=m;
        cout<<n-3+count<<endl;
        fo1(i,n-3){
            cout<<i+2<<" "<<n<<endl;
        }
        fo(i,count)cout<<n<<" "<<2<<endl;
        }
    else{
        m=n;
        while(n>1){
            n= (n+15)/16;
            count++;
        }
        n=m;
        cout<<n-5+4+count<<endl;
        fo1(i,n-3){
            if(i!=14 && i!=2)cout<<i+2<<" "<<n<<endl;
        }
        fo(i,count)cout<<n<<" "<<16<<endl;
        fo(i,2)cout<<16<<" "<<4<<endl;
        fo(i,2)cout<<4<<" "<<2<<endl;
    }





        
    }
    cerr << "Time : " << 1000 * ((double)clock()) / (double)CLOCKS_PER_SEC << "ms\n";
    return 0;
}