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: Accepted)

class Combination
  MOD = 1_000_000_007
  def initialize
    @fact = (1..2000000).to_a
    @fact.each_with_index do |e, idx|
      @fact[idx] = @fact[idx] * @fact[idx-1] % MOD if idx > 0
    end
  end

  def comb(n, r)
    return 0 if not (0 <= r and r <= n)
    return 1 if r == 0 or r == n
    fact(n) * inverse( fact(r) * fact(n-r) % MOD ) % MOD
  end

  def fact(n)
    @fact[n-1]
  end

  def inverse(x)
    pow(x, MOD-2)
  end

  def pow(x, n)
    ans = 1
    while n > 0
      ans = ans * x % MOD if n.odd?
      x = x * x % MOD
      n >>= 1
    end
    ans
  end
end


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

if (2 * Y - X) % 3 == 0 && (2 * X - Y) % 3 == 0
  a = (2 * Y - X) / 3
  b = (2 * X - Y) / 3
  puts Combination.new.comb(a + b, a)
else
  puts 0
end