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 PyPy3 (2.4.0) (judge verdict: Wrong Answer)

r1, c1, r2, c2 = map(int, input().split())

class factorial:
  def __init__(self, n, p=10**9 + 7):
    self.__p  = p
    self.__fac  = [0] * (n+1)
    self.__inv  = [0] * (n+1)
    self.__finv = [0] * (n+1)

    self.__fac[0] = self.__fac[1] = 1
    self.__inv[1] = 1
    self.__finv[0] = self.__finv[1] = 1

    for i in range(2, n+1):
      self.__fac[i] = self.__fac[i-1] * i % p
      self.__inv[i] = p - self.__inv[p%i] * (p // i) % p
      self.__finv[i] = self.__finv[i-1] * self.__inv[i] % p
  
  def fact(self, n):
    return self.__fac[n]
  
  def perm(self, n, k):
    return self.__fac[n] * self.__finv[n-k] % self.__p
  
  def comb(self, n, k):
    return self.perm(n, k) * self.__finv[k] % self.__p

f = factorial(r2+c2+2)
print(f.comb(r2+c2+2, r2+1) + f.comb(r1+c1, r1) - f.comb(r1+c2+1, r1) - f.comb(c1+r2+1, c1))