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 ll long long int
#define endl "\n"
#define io ios_base::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define mod 1000000007
#define pb push_back
#define vl vector<ll>
#define vs vector<string>
#define vvl vector<vector<ll>>
#define vb vector<bool>
#define vvb vector<vector<bool>>
#define ull unordered_map<ll, ll>
#define usl unordered_map<string, ll>
#define um unordered_map
#define pll pair<ll, ll>
using namespace std;
void solve();
string conc(ll a, ll b) { return to_string(a) + "|" + to_string(b); }
ll rev(ll x){ ll ans = 0; while (x != 0) { ans = ans * 10 + (x % 10); x /= 10; } return ans; }
ll numdig(ll x) { ll ans = 0; while (x != 0) { ans++; x /= 10; } return ans; }
void printvec(vector<ll> &v) { for (ll i = 0; i < v.size(); i++) cout << v[i] << " "; cout << endl; }
void printmat(vector<vector<ll>> &mat) { for (ll i = 0; i < mat.size(); i++) { for (ll j = 0; j < mat[0].size(); j++) cout << mat[i][j] << " "; cout << endl; } }

void solve(){
    ll n;
    cin>>n;
    if(n==3) cout<<"2\n3 2\n3 2\n";
    else if(n==4) cout<<"3\n3 4\n4 2\n4 2\n";
    else if(n<=8){
        cout<<n<<endl;
        for(ll i=3; i<n; i++) cout<<i<<" "<<n<<endl;
        ll curr = n;
        while(curr!=1){
            curr = ceil(1.0*curr/2);
            cout<<n<<" 2\n";
        }
    }
    else{
        cout<<n+5<<endl;
        for(ll i=1; i<=n; i++){
            if(i!=1 && i!=2 && i!=4 && i!=8 && i!=n) cout<<i<<" "<<n<<endl;
        }
        for(ll i=0; i<6; i++) cout<<n<<" 8\n";
        cout<<"8 4\n8 4\n4 2\n4 2\n";
    }
}

int main(void) {
    io
    ll t;
    t = 1;
    cin >> t;
    while (t--) solve();
    return 0;
}