/*
    題意：給出一條直線上的點(diǎn)，每個(gè)點(diǎn)xi有值vi  每一對(duì)(i,j)的值為：dis(ti,j)*max(vi,vj)
           現(xiàn)求所有C(N,2)對(duì)點(diǎn)的所有值   O(n^2)會(huì)Tle
    先按照V排序！
    用兩個(gè)一維樹(shù)狀數(shù)組，分別記錄小于xi的點(diǎn)的個(gè)數(shù)pre_cnt以及小于xi點(diǎn)的距離之和pre_dist
    對(duì)于i之前的點(diǎn)xj，分xj<xi和xj>xi計(jì)算：
    ans+=(long long)v*(pre_cnt*x-pre_dist + total_dist-pre_dist-(i-pre_cnt)*x);
    total_dist為i-1個(gè)點(diǎn)的x坐標(biāo)之和
*/
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

const int MAXN = 20000;

struct Cow
{
    int v,x;
    bool operator<(const Cow &c)const
    {
        return v<c.v;
    }
}cows[MAXN+5];

int cnt_C[MAXN+5],dist_C[MAXN+5];
int N;

int lowbit(int x){return x&(-x);}

void insert(int *a,int p,int x)
{
    while(p<=MAXN)
    {
        a[p]+=x;
        p+=lowbit(p);
    }
}

int sum(int *a,int p)
{
    int ans = 0;
    while(p>0)
    {
        ans+=a[p];
        p-=lowbit(p);
    }
    return ans;
}

int main()
{
    while(~scanf("%d",&N))
    {
        for(int i=0;i<N;i++)
            scanf("%d%d",&cows[i].v,&cows[i].x);
        sort(cows,cows+N);
        long long ans = 0;
        int total_dist = 0;
        for(int i=0;i<N;i++)
        {
            int x = cows[i].x,v=cows[i].v;
            int pre_dist = sum(dist_C,x),pre_cnt = sum(cnt_C,x);//no two in the same pos
            ans+=(long long)v*(pre_cnt*x-pre_dist + total_dist-pre_dist-(i-pre_cnt)*x);//i從0開(kāi)始
            insert(dist_C,x,x);
            insert(cnt_C,x,1);
            total_dist+=x;
        }
        printf("%lld\n",ans);
    }
    return 0;
}


hdu 3015 跟這個(gè)類(lèi)似，不過(guò)要先離散化