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: Wrong Answer)

X, Y = gets.split.map(&:to_i)
M = 10**9 + 7

require 'matrix'

def mpow(x, n)
  ret = 1
  while n > 0
    if n.even?
      x = x * x % M
      n >>= 1
    else
      ret = ret * x % M
      n -= 1
    end
  end
  ret
end

f = ->(a, b){(a .. b).inject(1){|r, x| r * x % M}}
a, b = Matrix[[2, 1], [1, 2]].lup.solve([X, Y]).to_a
unless a.denominator == 1 && b.denominator == 1
  puts 0
else
  a, b = a.to_i, b.to_i
  n = a + b
  k = [a, b].min
  puts f[n - k + 1, n] * mpow(f[1, k], M - 2) % M
end