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: Wrong Answer)

X,Y = list(map(int,input().split(" ")))



MOD = 10 ** 9 + 7

def inv_fact(a,mod = MOD): #あるmod下における a ! の逆元を求める関数
    ans = 1
    for i in range(1,a + 1):
        ans *= pow(i, MOD - 2 ,MOD)
        ans %= MOD
    return ans
NOTFOUND = True
for a in range(X):
    # a → 上に2,横に1の移動
    # b → 上に1,横に2の移動
    b = (X - a) //2
    if 2 * a + b == Y and (X - a) % 2 == 0: #上手くハマるbが見つかったら
        NOTFOUND = False
        break
if NOTFOUND: #
    print(0)
else:
    # (a + b) ! *(a! )-1 * (b!) -1
    a_inv =inv_fact(a)
    b_inv = inv_fact(b)
    ans = 1
    for i in range(1,a + b + 1):
        ans *= i
        ans %= MOD
    mul = a_inv * b_inv %MOD
    ans *=(mul)
    ans %= MOD
    print(ans)