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 PyPy3 (2.4.0) (judge verdict: Accepted)

mod = 1000000007
import sys
sys.setrecursionlimit(1000000)
x, y = map(int, input().split())
a = ((2*x-y)//3)
b = ((2*y-x)//3)
if (x+y) % 3 != 0 or a < 0 or b < 0:
    print(0)
else:
    n = (a + b)%mod
    kaijo = [1, 1]
    for i in range(2, n+1):
        kaijo.append(i*kaijo[i-1]%mod)
    def mul(a, b):
        return ((a % mod) * (b % mod)) % mod
    def power(x, y):
        if y == 0:
            return 1
        elif y == 1:
            return x % mod
        elif y % 2 == 0:
            return power(x, y // 2) ** 2 % mod
        else:
            return power(x, y // 2) ** 2 * x % mod
    def div(a, b):
        return mul(a, power(b, mod - 2))
    n_a = div(kaijo[n], kaijo[a])
    ans = div(n_a, kaijo[n-a])
    print(ans)