/*
f[i]=f[j]+a[j+1].y*a[i].x;
f[j]=-a[j+1].y*a[i].x + f[i];
yj=f[j]
xj=-a[j+1].y 
k=a[i].x
b=f[i]
*/ 
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=51100;
struct node{ LL x,y;}a[N];
bool cmp(node n1,node n2){if( n1.x!=n2.x) return n1.x<n2.x;else return n1.y<n2.y;}
LL  f[N];
int q[N];
double X(int j){ return -1.0*a[j+1].y;}
double Y(int j){ return 1.0*f[j];} 
double slop(int j1,int j2){return (X(j2)==X(j1))?1e18*( Y(j2)-Y(j1) ):( Y(j2)-Y(j1) )/( X(j2)-X(j1) );}
int main()
{
    int n;scanf("%lld%lld", &n);
    for(int i=1;i<=n;i++)scanf("%lld%lld", &a[i].x, &a[i].y);
    sort(a+1,a+n+1,cmp);
    int tn=1;
    for(int i=2;i<=n;i++)
    {
        while(tn>0 && a[tn].y<=a[i].y)tn--;
        a[++tn]=a[i];
    }
    int l=1,r=1;q[1]=0;n=tn;
    for(int i=1;i<=n;i++)
    {
        while(l<r && slop(q[l],q[l+1])<a[i].x)l++;
        f[i]= f[q[l]]+ a[q[l]+1].y*a[i].x ;
        while(l<r && slop(q[r-1],q[r])>slop(q[r],i) )r--;
        q[++r]=i;
    }
    printf("%lld\n",f[n]);
    return 0;
}

/*
f[i]=f[j]+a[j+1].y*a[i].x;
f[j]=-a[j+1].y*a[i].x + f[i];
yj=f[j]
xj=-a[j+1].y 
k=a[i].x
b=f[i]
*/ 
#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=51100;
struct node{ LL x,y;}a[N];
bool cmp(node n1,node n2){if( n1.x!=n2.x) return n1.x<n2.x;else return n1.y<n2.y;}
LL  f[N];
int q[N];
double X(int j){ return -1.0*a[j+1].y;}
double Y(int j){ return 1.0*f[j];} 
double slop(int j1,int j2){return (X(j2)==X(j1))?1e18*( Y(j2)-Y(j1) ):( Y(j2)-Y(j1) )/( X(j2)-X(j1) )  ;}
int main()
{
    int n;scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%lld%lld",&a[i].x,&a[i].y);
    sort(a+1,a+n+1,cmp);//每块长方形排序：先考虑X从小到大，如果X相等再考虑Y从小到大 
    int tn=1;
    for(int i=2;i<=n;i++)//经常这样优化数据 
    {
        while(tn>0 && a[tn].y<=a[i].y)tn--;
        a[++tn]=a[i];
    }
    int l=1,r=1;q[1]=0;n=tn;
    for(int i=1;i<=n;i++)
    {
        while(l<r && slop(q[l],q[l+1])<a[i].x)l++;
        f[i]= f[q[l]]+ a[q[l]+1].y*a[i].x ;
        while(l<r && slop(q[r-1],q[r])>slop(q[r],i) )r--;
        q[++r]=i;
    }
    printf("%lld\n",f[n]);
    return 0;
}