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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.math.BigInteger;
import java.util.Scanner;

public class Main
{

	static public BigInteger mod = new BigInteger("1000000007");
	public static void main(String[] args)
	{
		Scanner scan = new Scanner(System.in);
		int r1 = scan.nextInt();
		int c1 = scan.nextInt();
		int r2 = scan.nextInt();
		int c2 = scan.nextInt();

		BigInteger ans = BigInteger.ZERO;
		BigInteger f1 = BigInteger.ZERO;
		for (int i=c1; i<c2+1;i++) {
			if (ans == BigInteger.ZERO) {
				f1 = f(c1, r1);
			} else {
				f1 =  f1.multiply(BigInteger.valueOf((long)(r1+i))).divide(BigInteger.valueOf((long)i));
			}
			ans = ans.add(f1);
			BigInteger f2 = f1;
			for (int j=r1+1;j<r2+1;j++) {
				f2 = f2.multiply(BigInteger.valueOf((long)(j+i))).divide(BigInteger.valueOf((long)j));
				ans = ans.add(f2);
			}
		}
		ans = ans.remainder(mod);
		System.out.println(ans);
	}

	static public BigInteger f(int r, int c) {

		/*
		if (c+r == 2 ) {
			BigInteger ans = new BigInteger("2");
			return ans;
		}
*/
		if (c > r) {
			int tmp = c;
			c = r;
			r = tmp;
		}

		BigInteger rcf = factorial_m(c,r);
		BigInteger rf = factorial(r);
		BigInteger ans = rcf.divide(rf);
		return ans;
	}
	static public BigInteger factorial(int n) {
	    BigInteger result = BigInteger.ONE;
	    for (int i = 2; i <= n; i++)
	        result = result.multiply(BigInteger.valueOf(i));
	    return result;
	}
	static public BigInteger factorial_m(int c, int r) {
	    BigInteger result = BigInteger.ONE;
	    for (int i = c+1; i <= c+r; i++)
	        result = result.multiply(BigInteger.valueOf(i));
	    return result;
	}

}