//貌似本题与旋转卡壳没有太大的关系
//个人觉得致力不错：http://www.cnblogs.com/mjy0724/p/4423905.html
#include<cstdio>#include<cmath>#include<algorithm>using namespace std;const int N=1e5+5;const double eps=1e-10;struct Vector{    double x,y;    Vector(double x=0,double y=0):x(x),y(y){}}p[N],ch[N];int n;Vector operator + (Vector A,Vector B){return Vector(A.x+B.x,A.y+B.y);}//向量加法Vector operator - (Vector A,Vector B){return Vector(A.x-B.x,A.y-B.y);}//向量减法Vector operator * (Vector A,double p){return Vector(A.x*p,A.y*p);}//向量乘以标量Vector operator / (Vector A,double p){return Vector(A.x/p,A.y/p);}//向量除以标量bool operator < (const Vector &a,const Vector &b){//点的坐标排序    return a.x==b.x?a.y<b.y:a.x<b.x;}int dcmp(double x){//三态函数，减少精度问题    if(fabs(x)<eps) return 0;    else return x<0?-1:1;//忘写return WA*1 }bool operator ==(const Vector &a,const Vector &b){//判断同一个点    return dcmp(a.x-b.x)==0&&dcmp(a.y-b.y)==0;}double Dot(Vector A,Vector B){return A.x*B.x+A.y*B.y;}//向量点积double Length(Vector A){return sqrt(Dot(A,A));} //向量长度，点积double Angle(Vector A,Vector B){return acos(Dot(A,B)/Length(A)/Length(B));}//向量转角，逆时针，点积double Cross(Vector A,Vector B){return A.x*B.y-A.y*B.x;}//向量叉积double Area2(Vector A,Vector B,Vector C){return Cross(B-A,C-A);}//三角形有向面积的两倍 //计算凸包，输入点数组p,不重复个数cnt,输出点数组ch。函数返回凸包顶点个数。//如果不希望在凸包的边上有输入点，把两个 <=改成 < int ConvexHull(){    sort(p,p+n);    int m=0;ch[0]=p[0];    for(int i=1;i<n;i++){        while(m>0&&Cross(ch[m]-ch[m-1],p[i]-ch[m-1])<=0) m--;        ch[++m]=p[i];    }    int k=m;    for(int i=n-2;i>=0;i--){        while(m>k&&Cross(ch[m]-ch[m-1],p[i]-ch[m-1])<=0) m--;        ch[++m]=p[i];    }    return m+=(n==1);}double maxarea(int n,Vector *p){    if(n<=2) return 0.0;    if(n==3) return fabs(Area2(p[0],p[1],p[2]));    static int nxt[N];    nxt[n-1]=0;    for(int i=0;i<n-1;i++) nxt[i]=i+1;    double rt=0.0;    for(int i=0;i<n;i++){        for(int j=nxt[nxt[i]],a=nxt[i],b=nxt[j];nxt[j]!=i;j=nxt[j]){            while(Area2(p[i],p[a],p[j])<Area2(p[i],p[nxt[a]],p[j])) a=nxt[a];            while(Area2(p[j],p[b],p[i])<Area2(p[j],p[nxt[b]],p[i])) b=nxt[b];            double ra=Area2(p[i],p[a],p[j]);            double rb=Area2(p[j],p[b],p[i]);            rt=max(rt,ra+rb);        }    }    return rt/2.0;}int main(){    scanf("%d",&n);    for(int i=0;i<n;i++) scanf("%lf%lf",&p[i].x,&p[i].y);    int c=ConvexHull();    printf("%.3lf\n",maxarea(c,ch));    return 0;}