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 Ruby (2.3.3) (judge verdict: Time Limit Exceeded)

MAX = 10000000 #二項係数n, k の n のsaidaiti
MOD = 1000000007

def mod_power(n, p, m)
  #フェルマーの小定理
  return 1 if p == 0
  if p % 2 == 0
    t = mod_power(n, p / 2, m)
    return t * t % m
  else
    return n * mod_power(n, p - 1, m) % m
  end
end

def cominit()
  #階乗と階乗の逆元を用意する
  @fac = []
  @fac[0] = 1
  @fac[1] = 1
  @finv = []
  @finv[0] = 1
  @finv[1] = 1

  2.upto(MAX) do |i|
    @fac[i] = @fac[i-1] * i % MOD
  end

  @finv[MAX] = mod_power(@fac[MAX], MOD - 2, MOD)
  (MAX-1).downto(2) do |i|
    @finv[i] = @finv[i+1] * (i + 1) % MOD
  end
end

def comb(n, r)
  return @fac[n] * @finv[r] * @finv[n-r] % MOD
end

def main
  r1, c1, r2, c2 = gets.chomp.split.map(&:to_i)
  ans = 0

  cominit

  (c1..c2).each do |i|
    ans -= comb(r1 + i, i + 1)
    ans += comb(r2 + i + 1, i + 1)
  end

  puts ans % MOD
end

main