Problem:
Score : 600 points

Problem Statement

Snuke is standing on a two-dimensional plane. In one operation, he can move by 1 in the positive x-direction, or move by 1 in the positive y-direction.

Let us define a function f(r, c) as follows:

f(r,c) := (The number of paths from the point (0, 0) to the point (r, c) that Snuke can trace by repeating the operation above)

Given are integers r_1, r_2, c_1, and c_2.
Find the sum of f(i, j) over all pair of integers (i, j) such that r_1 ≤ i ≤ r_2 and c_1 ≤ j ≤ c_2, and compute this value modulo (10^9+7).

Constraints

1 ≤ r_1 ≤ r_2 ≤ 10^6

1 ≤ c_1 ≤ c_2 ≤ 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

r_1 c_1 r_2 c_2

Output

Print the sum of f(i, j) modulo (10^9+7).

Sample Input 1

1 1 2 2

Sample Output 1

14

For example, there are two paths from the point (0, 0) to the point (1, 1): (0,0) → (0,1) → (1,1) and (0,0) → (1,0) → (1,1), so f(1,1)=2.

Similarly, f(1,2)=3, f(2,1)=3, and f(2,2)=6. Thus, the sum is 14.

Sample Input 2

314 159 2653 589

Sample Output 2

602215194

Solution in C (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>

#define mod 1000000007
#define ll long long

int r1,c1,r2,c2;
int st,la;
ll ans=0;

ll pow2(ll a,ll n){
  /*
  if(n==1) return a;
  if(n%2==0){
    return pow2(a*a%mod,n/2)%mod;}
  else{
    return a*pow2(a-1,n)%mod;}
  */
  ll x=1;
  for(a%=mod;n;n/=2) n&1 ? x=x*a%mod:0,a=a*a%mod;
  return x;
}

ll com(int n,int k){
  ll ans=1;
  for(int i=0;i<k;i++){
    ans = ans*(n-i)%mod*pow2(k-i,mod-2)%mod;}
  return ans;}

ll g(int r,int c){
  return com(r+1+c+1,r+1);
}

int main(){
  scanf("%d %d %d %d",&r1,&c1,&r2,&c2);

  ll ans = g(r2,c2)-g(r2,c1-1)-g(r1-1,c2)+g(r1-1,c1-1);
  ans = (ans+mod)%mod;
  printf("%lld\n",ans);
}