Problem:
Score : 400 points

Problem Statement

There is a knight - the chess piece - at the origin (0, 0) of a two-dimensional grid.

When the knight is at the square (i, j), it can be moved to either (i+1,j+2) or (i+2, j+1).

In how many ways can the knight reach the square (X, Y)?

Find the number of ways modulo 10^9 + 7.

Constraints

1 \leq X \leq 10^6

1 \leq Y \leq 10^6

All values in input are integers.

Input

Input is given from Standard Input in the following format:

X Y

Output

Print the number of ways for the knight to reach (X, Y) from (0, 0), modulo 10^9 + 7.

Sample Input 1

3 3

Sample Output 1

2

There are two ways: (0,0) \to (1,2) \to (3,3) and (0,0) \to (2,1) \to (3,3).

Sample Input 2

2 2

Sample Output 2

0

The knight cannot reach (2,2).

Sample Input 3

999999 999999

Sample Output 3

151840682

Print the number of ways modulo 10^9 + 7.

Solution in Ruby (2.3.3) (judge verdict: Runtime Error)

require 'openssl'

X, Y = gets.split.map(&:to_i)

MOD = OpenSSL::BN.new(1000000007)

def factorial(n)
  return OpenSSL::BN.new(1) if n.zero?
  ret = (1..n).inject{|result, index|
    result = (result * index) % MOD
  }
  OpenSSL::BN.new(ret)
end

def able_to?(x, y)
  return true if x % 3 == 1 && y % 3 == 2
  return true if x % 3 == 2 && y % 3 == 1
  return true if x % 3 == 0 && y % 3 == 0
  return false
end

if able_to?(X, Y) then
  step_x = OpenSSL::BN.new(X / 2)
  step_y = OpenSSL::BN.new((Y - step_x) / 2)

  fac_xy = factorial(step_x + step_y)
  fac_x = factorial(step_x)
  fac_y = factorial(step_y)

  # factorial(step_x + step_y) / (factorial(step_x) * factorial(step_y))
  # ans = fac_xy * fac_x.mod_inverse(MOD) % MOD * fac_y.mod_inverse(MOD) % MOD
  ans = fac_xy * (fac_x * fac_y % MOD).mod_inverse(MOD) % MOD
else
  ans = 0
end

puts ans