#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<<" "; } cout<<ans<<' '; }