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 (2.7.1) (judge verdict: Wrong Answer)

    x,y = gets.chomp.split(" ").map(&:to_i)
 
    a = (2 * y) - x
    b = (2 * x) - y
    
	MOD = (10 ** 9) + 7
 
    
    if a < 0 or b < 0 or a % 3 != 0 or b % 3 != 0
      puts 0
      exit
    end
      
    def kaijo(a, b)
      result = 1
      ([a,b].min..[a,b].max).each {|v| result = (result * v) % MOD}
      return result % MOD
    end
 
    def joho_gyakugen(a, b)
      #modを法とする逆元を計算
      result = 1
      ([a,b].min..[a,b].max).each {|v| result = (result * v) % MOD}
      return ruijo(result, MOD-2)
    end
    
    def ruijo(tei, bekisu)
      if bekisu == 0
        return 1
      elsif bekisu % 2 == 0
        result = ruijo((tei * tei) % MOD, bekisu / 2)
      else
        result = tei * ruijo((tei * tei) % MOD, bekisu / 2)
      end
      return result % MOD
    end
 
    n = [a, b].min / 3
    m = [a, b].max / 3
    total = n + m
    result = (kaijo(m+1, total) * joho_gyakugen(1, n))
    puts result % MOD