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 Python (3.4.3) (judge verdict: Accepted)

# https://atcoder.jp/contests/abc145/tasks/abc145_d
from math import factorial
x, y = map(int, input().split())
mod = 10 ** 9 + 7

# two_steps = x // 2
# one_steps = x % 2
# ans = 0
# while two_steps:
#     if one_steps * 2 + two_steps * 1 == y:
#         ans += (factorial(one_steps + two_steps) * (pow(factorial(one_steps), mod - 2, mod) * pow(factorial(two_steps), mod - 2, mod))) % mod
#     two_steps -= 1
#     one_steps += 2
# print(ans)

MAX = 10 ** 6 + 1

def com_init(MAX, mod):
    fac = [0] * MAX
    finv = [0] * MAX
    inv = [0] * MAX
    fac[0] = fac[1] = 1
    finv[0] = finv[1] = 1
    inv[1] = 1
    for i in range(2, MAX):
        fac[i] = fac[i - 1] * i % mod
        inv[i] = mod - inv[mod % i] * (mod // i) % mod
        finv[i] = finv[i - 1] * inv[i] % mod
    return fac, finv

fac, finv = com_init(MAX, mod)

def nCr(n, r):
    if n < r:
        return 0
    elif n < 0 or r < 0:
        return 0
    return fac[n] * (finv[r] * finv[n - r] % mod) % mod 

if (x + y) % 3 == 0:
    a = (-x + 2 * y) // 3
    b = (2 * x - y) // 3
    ans = nCr(a + b, a)
    # ans = (factorial(a + b) * (pow(factorial(a), mod - 2, mod) * pow(factorial(b), mod - 2, mod))) % mod
else:
    ans = 0

print(ans)