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 ll long long int
#define MOD 1000000007 
#define oo 1000000000000000000
#define forr(i,n) for(ll i=0;i<n;i++)
#define fastio ios_base::sync_with_stdio(false); cin.tie(0) ; cout.tie(0);
#define all(x) x.begin(),x.end()
#define eb emplace_back
#define mem(a,v) memset(a,v,sizeof(a))
#define pb push_back
#define popcount(x) __builtin_popcount(x)

using namespace std;

void __print(int x) {cerr << x;}
void __print(long x) {cerr << x;}
void __print(long long x) {cerr << x;}
void __print(unsigned x) {cerr << x;}
void __print(unsigned long x) {cerr << x;}
void __print(unsigned long long x) {cerr << x;}
void __print(float x) {cerr << x;}
void __print(double x) {cerr << x;}
void __print(long double x) {cerr << x;}
void __print(char x) {cerr << '\'' << x << '\'';}
void __print(const char *x) {cerr << '\"' << x << '\"';}
void __print(const string &x) {cerr << '\"' << x << '\"';}
void __print(bool x) {cerr << (x ? "true" : "false");}

template<typename T, typename V>
void __print(const pair<T, V> &x) {cerr << '{'; __print(x.first); cerr << ','; __print(x.second); cerr << '}';}
template<typename T>
void __print(const T &x) {int f = 0; cerr << '{'; for (auto &i: x) cerr << (f++ ? "," : ""), __print(i); cerr << "}";}
void _print() {cerr << "]\n";}
template <typename T, typename... V>
void _print(T t, V... v) {__print(t); if (sizeof...(v)) cerr << ", "; _print(v...);}
#ifndef ONLINE_JUDGE
#define debug(x...) cerr << "[" << #x << "] = ["; _print(x)
#else
#define debug(x...)
#endif

struct custom_hash {
    static uint64_t splitmix64(uint64_t x) {
        x += 0x9e3779b97f4a7c15;
        x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
        x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
        return x ^ (x >> 31);
    }
    size_t operator()(uint64_t x) const {
        static const uint64_t FIXED_RANDOM = chrono::steady_clock::now().time_since_epoch().count();
        return splitmix64(x + FIXED_RANDOM);
    }
};

ll add(ll a, ll b,ll p=MOD) { a%=p; b%=p; return (a+b + p)%p;}
ll mul(ll a, ll b,ll p=MOD) { a%=p; b%=p; return (a*b + p)%p;} // __int128
ll power(ll x,ll n,ll p=MOD){ if(x==0) return 0; if(n==0 || x==1) return 1LL;
    ll r = (power(x,n/2,p))%p; if(n&1) return mul(mul(r,r,p) , x,p); else return mul(r,r,p);
}
ll inv(ll x){return power(x,MOD-2);}


pair<int,int> dx[4] = { { -1,0 } , {1,0} , {0,-1} , {0,1} };


// Now we can simply define our unordered_map or our gp_hash_table as follows:
// unordered_map<long long, int, custom_hash> safe_map;
// gp_hash_table<long long, int, custom_hash> safe_hash_table;

/* ------------------------------------------ Code ----------------------------------------------- */





void __sol(){
    int n; cin >> n;
    int cnt = 0;
    if(n<=32){
        vector<pair<int,int>> v;
        for(int i=1;i<n;i++){
            if(i!=2) v.eb(i,n),cnt++;
        }
        int last = n;
        while(n!=1){
            v.eb(last,2);
            n=n/2 + n%2;
            cnt++;
        }
        cout << v.size()<<"\n";
        for(auto &i: v) cout << i.first<<" "<<i.second<<"\n";
          //  cerr<<cnt <<"\n";
        return;
    }
    vector<pair<int,int>> v;
    int temp = n;
    int lg = 0;
    while(temp!=1){
        temp/=2;
        lg++;
    }
    temp = power(2,lg);
    for(int i=3;i<n;i++){
        if(i!=32){
            v.eb(i,n);
        }
    }
    int last = n;
    while(last!=1){
        v.eb(n,32);
        last=last/32 + (last%32!=0);
    }
    v.eb(32,2);
    v.eb(32,2);
    v.eb(32,2);
    v.eb(32,2);
    v.eb(32,2);
    cout << v.size()<<"\n";
    cerr<<v.size()<<"\n";
    for(auto &i: v) cout <<i.first<<" "<< i.second<<"\n";

    return;
}


int main(){
    #ifndef ONLINE_JUDGE
    freopen("input.txt","r",stdin);
    freopen("output.txt","w",stdout);
    #endif
    clock_t time_req;  
    time_req = clock(); 
    fastio
    ll tc=1;   cin >> tc;
    while(tc--) __sol();
    time_req = clock()- time_req; 
    cerr<< "Passed in: "<< (float)time_req/CLOCKS_PER_SEC << " seconds" << endl; 
    return 0;
}