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: Time Limit Exceeded)

#include<stdio.h>
int f(int r,int c);

int main(void){
  int r1,r2,c1,c2;
  int ans=0;
  scanf("%d %d %d %d\n",&r1,&c1,&r2,&c2 );
  for(int i=r1; i<=r2;i++){
    for(int j=c1; j<=c2;j++){
      ans += f(i,j);
    }
  }
  printf("%d\n",ans );
  return 0;
}

int f(int r,int c){
  int ans;
  //if(dp[r][c] != -1) return dp[r][c];

  if(c==0 && r==0) ans = 1;
  else if(c == 0 && r != 0) ans = f(r-1,c);
  else if(c != 0 && r == 0) ans = f(r,c-1);
  else ans = f(r-1,c) + f(r,c-1);
  return /*dp[r][c] = */ans;
}