Problem:
Score : 1400 points

Problem Statement

Snuke is having another barbeque party.

This time, he will make one serving of Skewer Meal.

He has a stock of N Skewer Meal Packs. The i-th Skewer Meal Pack contains one skewer, A_i pieces of beef and B_i pieces of green pepper.
All skewers in these packs are different and distinguishable, while all pieces of beef and all pieces of green pepper are, respectively, indistinguishable.

To make a Skewer Meal, he chooses two of his Skewer Meal Packs, and takes out all of the contents from the chosen packs, that is, two skewers and some pieces of beef or green pepper.
(Remaining Skewer Meal Packs will not be used.)
Then, all those pieces of food are threaded onto both skewers, one by one, in any order.

(See the image in the Sample section for better understanding.)

In how many different ways can he make a Skewer Meal? Two ways of making a Skewer Meal is different if and only if the sets of the used skewers are different, or the orders of the pieces of food are different.
Since this number can be extremely large, find it modulo 10^9+7.

Constraints

2≦N≦200,000

1≦A_i≦2000, 1≦B_i≦2000

Input

The input is given from Standard Input in the following format:

N
A_1 B_1
A_2 B_2
:
A_N B_N

Output

Print the number of the different ways Snuke can make a serving of Skewer Meal, modulo 10^9+7.

Sample Input 1

3
1 1
1 1
2 1

Sample Output 1

26

The 26 ways of making a Skewer Meal are shown below.
Gray bars represent skewers, each with a number denoting the Skewer Meal Set that contained the skewer.
Brown and green rectangles represent pieces of beef and green pepper, respectively.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
#define ll int
#define llu unsigned long long
#define ld double
#define llu unsigned long long
#define rep(i,x,y)  for(int i=int(x);i<int(y);++i)
#define For(i,x,y)  for(int i=int(x);i<=int(y);++i)
#define FOr(i,x,y)  for(int i=int(x);i>=int(y);--i)
#define pi acos(-1) 
#define mk make_pair
#define pa pair<ll,ll>
#define lf else if
#define IL inline
#define max(x,y)    ((x)<(y)?(y):(x))
#define min(x,y)    ((x)<(y)?(x):(y))
#define sqr(x)      ((x)*(x))
#define Mul(x,y)    ((x)=1LL*(x)*(y)%mod)
#define Add(x,y)    ((x)=((x)+(y))%mod)
#define Max(x,y)    ((x)=((x)<(y)?(y):(x)))
#define Min(x,y)    ((x)=((x)>(y)?(y):(x)))
#define E(x)        return writeln(x),0
#define p(x)        printf("~%lld~\n",LL(x))
#define pp(x,y)     printf("~~%lld %lld~~\n",LL(x),LL(y))
#define LL			(long long)
#define ppp(x,y,z)  printf("~~~%lld %lld %lld~~~\n",LL(x),LL(y),LL(z))
#define pppp(a,b,c,d)	printf("~~~%lld %lld %lld %lld\n",LL(a),LL(b),LL(c),LL(d))
#define f_in(x)     freopen(x".in","r",stdin)
#define f_out(x)    freopen(x".out","w",stdout)
#define open(x)     f_in(x),f_out(x)
#define fi first
#define se second
#define GuYue		puts("\nGuYueNa________________________________________________________________________________")
typedef complex<double> E;
namespace SHENZHEBEI{
#ifdef LOCAL
    struct _{_(){freopen("cf.in","r",stdin);}}_; 
#endif 
#define NEG 1
    static const int GYN=2333333;
    char SZB[GYN],*S=SZB,*T=SZB;
    inline char gc(){   if (S==T){  T=(S=SZB)+fread(SZB,1,GYN,stdin); if (S==T) return '\n';  }   return *S++;    }
#if NEG
    inline ll read(){    ll x=0,g=1; char ch=gc();   for (;!isdigit(ch);ch=gc()) if (ch=='-') g=-1;  for (;isdigit(ch);ch=gc())  x=x*10-48+ch;   return x*g; }
    inline void write(ll x){    if (x<0) putchar('-'),x=-x;  if (x>=10)   write(x/10);    putchar(x%10+'0');  }
#else
    inline ll read(){    ll x=0; char ch=gc();   for (;!isdigit(ch);ch=gc());    for (;isdigit(ch);ch=gc())  x=x*10-48+ch;   return x;   }
    inline void write(ll x){    if (x>=10)   write(x/10);    putchar(x%10+'0');  }
#endif
    inline char readchar(){ char ch=gc();   for(;isspace(ch);ch=gc());  return ch;  }
    inline ll readstr(char *s){ char ch=gc();   int cur=0;  for(;isspace(ch);ch=gc());      for(;!isspace(ch);ch=gc())  s[cur++]=ch;    s[cur]='\0';    return cur; }
	void Print(long long *a,int s,int t){For(i,s,t)printf("%lld ",a[i]);}
	void Print(int *a,int s,int t){For(i,s,t)printf("%d ",a[i]);}
	void Print(char *a,int s,int t){For(i,s,t)putchar(a[i]);}
	void writeln(ll x){write(x);puts("");}
}using namespace SHENZHEBEI;
const ll N=200100,mod=1e9+7;
ll mp[4100][4100],a[N],b[N],n,cogito=2000,answ,c[4100][4100];
int main(){
	For(i,0,cogito*2)For(j,0,cogito*2){
		if (!i&&!j)	c[i][j]=1;
		lf (!i)		c[i][j]=c[i][j-1];
		lf (!j)		c[i][j]=c[i-1][j];
		else		c[i][j]=(c[i-1][j]+c[i][j-1]);
	}
	n=read();
	For(i,1,n)a[i]=read(),b[i]=read(),mp[-a[i]+cogito][-b[i]+cogito]++;
	For(i,0,2*cogito)For(j,0,2*cogito){
		if(i)Add(mp[i][j],mp[i-1][j]);
		if(j)Add(mp[i][j],mp[i][j-1]);	
	}
	For(i,1,n)Add(answ,mp[a[i]+cogito][b[i]+cogito]),Add(answ,-c[a[i]*2][b[i]*2]);
	writeln((1ll*answ*(mod+1)/2%mod+mod)%mod);
}