#include<bits/stdc++.h>

#define ll long long

using namespace std;

const int N=1e6+10;

const ll INF=2e18;

ll n,f[N],x[N],p[N],c[N],q[N],l,r,sum[N],sump[N];//f[i][0/1]表示i及之后的工厂产品

inline int read()//i工厂建仓库的最小代价

{

    int x=0,ff=1;

    char ch=getchar();

    while(!isdigit(ch)) {if(ch=='-') ff=-1;ch=getchar();}

    while(isdigit(ch)) {x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}

    return x*ff;

}

inline double X(int i) {return sump[i];}

inline double Y(int i) {return f[i]+sum[i];}

inline double xie(int i,int j) {return (Y(i)-Y(j))/(X(i)-X(j));}

int main()

{

//    freopen("1.in","r",stdin);

    n=read();

    for(int i=1;i<=n;++i) x[i]=read(),p[i]=read(),c[i]=read();

    for(int i=1;i<=n;++i) sum[i]=sum[i-1]+x[i]*p[i],sump[i]=sump[i-1]+p[i];

    for(int i=1;i<=n;++i) f[i]=INF;

    f[0]=0;

    for(int i=1;i<=n;++i)

    {

        while(l<r&&xie(q[l],q[l+1])<=x[i]) ++l;

        int j=q[l];

        f[i]=f[j]+(sump[i-1]-sump[j])*x[i]-(sum[i-1]-sum[j])+c[i];

        while(l<r&&xie(i,q[r])<=xie(q[r],q[r-1])) --r;

        q[++r]=i;

        //for(int j=0;j<i;++j) f[i]=min(f[i],f[j]+(sump[i-1]-sump[j])*x[i]-(sum[i-1]-sum[j]));

        //f[i]+=c[i];

    }

    printf("%lld",f[n]);

    return 0;

}