#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define N 500005
struct point{
ll x,y;
point operator + (const point &k1) const{return (point){k1.x+x,k1.y+y};}
point operator - (const point &k1) const{return (point){x-k1.x,y-k1.y};}
};
ll cross(point k1,point k2){return k1.x*k2.y-k1.y*k2.x;}
point q[N];
int head,tail;
void pop(ll k){
while(tail>head && q[head+1].y-q[head].y<=k*(q[head+1].x-q[head].x))
head++;
}
void push(point k1){
while(tail>head && cross(q[tail]-k1,q[tail-1]-k1)>=0)tail--;
q[++tail]=k1;
}
ll n,dp[N],d[N],w[N],sumd[N],sumw[N],c[N];
int main(){
cin>>n;
for(int i=1;i<=n;i++)scanf("%lld%lld",&w[i],&d[i]);
for(int i=1;i<=n+1;i++)sumw[i]=sumw[i-1]+w[i];
for(int i=1;i<=n+1;i++)sumd[i]=sumd[i-1]+d[i-1];
for(int i=1;i<=n+1;i++)c[i]=c[i-1]+d[i-1]*sumw[i-1];//把所有树移到位置i的代价
ll ans=0x3f3f3f3f3f3f3f3f;
head=tail=1;
q[1]=(point){sumw[1],sumw[1]*sumd[1]};//第一个点已经在图上了
for(int i=2;i<=n;i++){
pop(sumd[i]);
dp[i]=q[head].y-q[head].x*sumd[i];
dp[i]+=c[n+1]-sumw[i]*sumd[n+1]+sumw[i]*sumd[i];
push((point){sumw[i],sumw[i]*sumd[i]});
ans=min(ans,dp[i]);
//cout<<dp[i]<<" "<<head<<" "<<tail<<"\n";
}
cout<<ans<<'\n';
}