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>
#include <stdlib.h>

#define P 1000000007

// a^x
long m(long a, int x){
	if(x==0) return 1;
	long b = m(a, x/2);
	if(x%2==0){
		b = b*b%P;
	}else{
		b = b*b%P*a%P;
	}
	return b;
}

// 1/a
long r(long a){
	return m(a, P-2);
}

// a!
long k(long a){
	long b=1;
	while(a>0){
		b=b*a%P;
		a--;
	}
	return b;
}

// (a+b)! / (a! b!)
long f(int a, int b){
	return r(k(a))*r(k(b))%P*k(a+b)%P;
}

int main(void){
	int r1, r2, c1, c2;
	scanf("%d%d%d%d\n", &r1, &c1, &r2, &c2);
	long ans = f(c2+1, r2+1) + f(c1, r1) - f(c2+1, r1) - f(c1, r2+1) + P*2;
	printf("%ld\n", ans%P);
}