#include<bits/stdc++.h>
#define int long long
#define ld long double 
#define pdd pair<ld,ld>
#define pii pair<int,int>
#define fi first
#define se second
#define pb emplace_back
using namespace std;
namespace IO{
	const int maxn=(1<<20);char *p1,*p2,buf[maxn];
	#define gc() (p1==p2&&(p2=(p1=buf)+fread(buf,1,maxn,stdin),p1==p2)?EOF:*p1++)
	int read(){
		int f=1,k=0;char c;
		while(!isdigit(c=gc()))if(c=='-')f=-1;
		while(isdigit(c))k=k*10+c-48,c=gc();
		return f*k;
	}
	void write(int k,char c){
		if(k<0){putchar('-');k=-k;}
		char st[21];int top=0;
		do{st[++top]=(k%10)|48,k/=10;}while(k);
		while(top)putchar(st[top--]);
		putchar(c);
	}
}using namespace IO;
const int N=1e4+10;
auto *it=new unsigned;
mt19937 rnd(time(0)^(*it));
const ld pi2=acos(-1)*2,M=1;
pdd P[N];int K,n,x,y,top;bool flag[N];
void add(ld x,ld y){P[++top]={x,y};}
int rd(){return (rnd()%4?1:-1)*(rnd()%K+1);}
namespace Andrew{
	#define x fi 
	#define y se
	int st[N],top;
	ld Cross(pdd a,pdd b,pdd c){
		return (b.x-a.x)*(c.y-a.y)-(b.y-a.y)*(c.x-a.x);
	}
	void solve(int n){
		sort(P+1,P+1+n);int ans=0;
		for(int i=1;i<=n;++i){
			while(top>1&&Cross(P[st[top-1]],P[st[top]],P[i])<=0)--top;
			st[++top]=i;
		}
		int t=top;
		for(int i=n;i>=1;--i){
			while(top>t&&Cross(P[st[top-1]],P[st[top]],P[i])<=0)--top;
			st[++top]=i;
		}
		for(int i=top;i>=1;--i)ans+=!flag[st[i]],flag[st[i]]=1;
		write(ans,'\n');memset(flag,0,sizeof flag);
		for(int i=top;i>=1;--i){
			if(flag[st[i]])continue;
			write(P[st[i]].x,' ');write(P[st[i]].y,'\n');
			flag[st[i]]=1;
		}
		write(n-ans,'\n');
		for(int i=1;i<=n;++i)
			if(!flag[i])write(P[i].x,' '),write(P[i].y,' '),write(rd(),'\n');	
	}
	#undef x
	#undef y 
}
signed main(){
	n=10000;K=1000;//n是题目中的点的总数，K是坐标的值域范围
	//n，K开太小有概率出现所有点共线的不和法情况   
	for(int i=1;i<=n;++i){
		x=rd(),y=rd();
		add(x,y);
	}
	Andrew::solve(n);
	return 0;
}