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>
#include <string>
#include <cmath>
#include <sstream>
#define FOR(i, a, b) for (int i = (a); i < (b); i++)
#define FORD(i, a, b) for(int i = (a); i >=(b); i--)
#define pb(a) push_back(a)
#define mp(a,b) make_pair(a,b)
typedef long long int ll;
using namespace std;
ll mod1=1000000007;
ll mod2=67280421310721;
ll mod3=998244353;
ll INF=1e18;

ll binpow1(ll a,ll b){
    ll res=1;
    while (b>0){
        if (b&1)
        res=res*a;
        a=a*a;
        b>>=1;
    }
    return res;
}

ll binpow2(ll a,ll b,ll m){
    a%=m;
    ll res=1;
    while (b>0){
        if (b & 1)
        res=res*a%m;
        a=a*a%m;
        b>>=1;
    }
    return res;
}

ll gcd(ll a,ll b){
    if (b==0)
        return a;
    else
        return gcd(b,a%b);
}

map<ll,ll> fact(ll n) {
    map<ll,ll> factorization;ll c=0;
    while (n % 2 == 0) {
        c++;
        n /= 2;
    }
    if(c>0)
    {
        factorization.insert({2,c});
    }c=0;
    for (ll d = 3; d * d <= n; d += 2) {
        while (n % d == 0) {
            c++;
            n /= d;
        }
        if(c>0)
    {
        factorization.insert({d,c});
    }c=0;
    }
    if (n > 1)
       {
           factorization.insert({n,1});
       }
    return factorization;
}

ll tree[200001]={0};ll ans2=0;

void update(ll l,ll r,ll i,ll k,ll a)
{
    if(i>r || i<l)
    {
        return ;
    }
    if(r==l && r==i)
    {
        tree[k]=a;
        return ;
    }
    ll mid=(l+r)/2;
    ll ans=0;
    update(l,mid,i,2*k,a);
    update(mid+1,r,i,2*k+1,a);
    tree[k]=gcd(tree[2*k],tree[2*k+1]);
}

void query(ll l,ll r,ll i,ll j,ll k)
{
    if(i>r || j<l)
    {
        return ;
    }
    else if(i<=l && j>=r)
    {
        ans2=gcd(ans2,tree[k]);
        return;
    }
    ll mid=(l+r)/2;
    query(l,mid,i,j,2*k);
    query(mid+1,r,i,j,2*k+1);
}

int main()
{ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	cout.tie(NULL);
    ll t;
    cin>>t;
    while(t--)
    {ll n;
    cin>>n;vector<pair<ll,ll>> v;
    if(n>100000)
    {
        FORD(i,n-1,101)
    {
        v.pb(make_pair(i,n));
    }
    v.pb(make_pair(n,100));
    FORD(i,100,11)
    {
        v.pb(make_pair(i,n));
    }


    if(n>10)
    {
        ll a=(n+99)/100;
        while(a!=1)
        {
            ll b=(a+9)/10;
            v.pb(make_pair(n,10));
            a=b;
        }
    }
    FORD(i,min(n-1,9LL),3)
    {
        v.pb(make_pair(i,min(10LL,n)));
    }
    ll a=min(n,10LL);
    while(a!=1)
        {
            ll b=(a+1)/2;
            v.pb(make_pair(min(n,10LL),2));
            a=b;
        }
    cout<<v.size()<<'\n';
    FOR(i,0,v.size())
    {
        cout<<v[i].first<<" "<<v[i].second<<'\n';
    }
    continue;
    }
    FORD(i,n-1,11)
    {
        v.pb(make_pair(i,n));
    }
    if(n>10)
    {
        ll a=n;
        while(a!=1)
        {
            ll b=(a+9)/10;
            v.pb(make_pair(n,10));
            a=b;
        }
    }
    FORD(i,min(n-1,9LL),3)
    {
        v.pb(make_pair(i,min(10LL,n)));
    }
    ll a=min(n,10LL);
    while(a!=1)
        {
            ll b=(a+1)/2;
            v.pb(make_pair(min(n,10LL),2));
            a=b;
        }
    cout<<v.size()<<'\n';
    FOR(i,0,v.size())
    {
        cout<<v[i].first<<" "<<v[i].second<<'\n';
    }
    }




	return 0;
}