首先我们看到这个题

发现符合条件的三角形一共有4种

然后我们对于这样一种情况算一下答案

化简后答案可得：

所以说我们对于每一个点，即可从上下左右四个方向算出来前缀和，然后统计答案

（代码自己写吧我写的很shit）

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<map>
#define MAXN 20005
#define int long long
using namespace std;
bool Test_MLE_start;
const int N=1e5+10;
const int mod=1e9+7;
int T=1,n,ans=0;
int prey[MAXN],prex[MAXN],cntx[MAXN],cnty[MAXN],lstx[MAXN],lsty[MAXN];
struct node{
	int x,y;
}a[N];
bool Test_MLE_end;
inline int reads(){
	char c=getchar();
	int sum=0,f=1;
	while(!isdigit(c)){
		if(c=='-') f=-1;
		c=getchar();
	}
	while(isdigit(c)){
		sum=(sum<<3)+(sum<<1)+(c-'0');
		c=getchar();
	}
	return sum*f;
}
inline void files(){
	freopen("std.in","r",stdin);
	freopen("std.out","w",stdout);
}
bool cmp1(node a,node b){return a.x==b.x?a.y<b.y:a.x<b.x;}
bool cmp2(node a,node b){return a.x==b.x?a.y>b.y:a.x<b.x;}
bool cmp3(node a,node b){return a.x==b.x?a.y<b.y:a.x>b.x;}
bool cmp4(node a,node b){return a.x==b.x?a.y>b.y:a.x>b.x;}
void clr(){
	memset(prex,0,sizeof(prex));
	memset(prey,0,sizeof(prey));
	memset(cntx,0,sizeof(cntx));
	memset(cnty,0,sizeof(cnty));
}
signed main(){
//	printf("%lf Mb\n",(&Test_MLE_end-&Test_MLE_start-1)/1024.0/1024.0);
//	files();
//	T=reads();
	while(T--){
		n=reads();
		for(int i=1;i<=n;i++){
			a[i].x=reads(),a[i].y=reads();
			a[i].x+=10000,a[i].y+=10000;
		}
		sort(a+1,a+n+1,cmp1);
		for(int i=1;i<=n;i++){
			int x=a[i].x,y=a[i].y;
			prex[x]=(prex[x]+abs(y-lstx[x])*cntx[x])%mod;
			cntx[x]++;
			lstx[x]=y;
			prey[y]=(prey[y]+abs(x-lsty[y])*cnty[y])%mod;
			cnty[y]++;
			lsty[y]=x;
			ans=(ans+prex[x]*prey[y])%mod;
		}
		sort(a+1,a+n+1,cmp2);
		clr();
		for(int i=1;i<=n;i++){
			int x=a[i].x,y=a[i].y;
			prex[x]=(prex[x]+abs(y-lstx[x])*cntx[x])%mod;
			cntx[x]++;
			lstx[x]=y;
			prey[y]=(prey[y]+abs(x-lsty[y])*cnty[y])%mod;
			cnty[y]++;
			lsty[y]=x;
			ans=(ans+prex[x]*prey[y])%mod;
		}
		sort(a+1,a+n+1,cmp3);
		clr();
		for(int i=1;i<=n;i++){
		int x=a[i].x,y=a[i].y;
			prex[x]=(prex[x]+abs(y-lstx[x])*cntx[x])%mod;
			cntx[x]++;
			lstx[x]=y;
			prey[y]=(prey[y]+abs(x-lsty[y])*cnty[y])%mod;
			cnty[y]++;
			lsty[y]=x;
			ans=(ans+prex[x]*prey[y])%mod;
		}
		sort(a+1,a+n+1,cmp4);
		clr();
		for(int i=1;i<=n;i++){
			int x=a[i].x,y=a[i].y;
			prex[x]=(prex[x]+abs(y-lstx[x])*cntx[x])%mod;
			cntx[x]++;
			lstx[x]=y;
			prey[y]=(prey[y]+abs(x-lsty[y])*cnty[y])%mod;
			cnty[y]++;
			lsty[y]=x;
			ans=(ans+prex[x]*prey[y])%mod;
		}
		printf("%lld\n",ans);
	}
	return 0;
}