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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;

LL const INF = 1000000007;


// mod. m での a の逆元 a^{-1} を計算する
LL inv(LL a, LL m) {
    LL b = m, u = 1, v = 0;
    while (b) {
        LL t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v ; swap(u, v);
    }
    u %= m;
    if (u < 0) u += m;
    return u;
}

LL fact(LL n) {
  if (n == 0) return 1;
  return n * fact(n - 1) % INF;
}

LL invfact(LL n) {
  if (n == 0) return 1;
  return inv(n, INF) * invfact(n - 1) % INF;
}

LL com(LL n, LL m) {
  return fact(n + m) * invfact(n) % INF * invfact(m) % INF;
}

int main() {
  
  LL X, Y;
  cin >> X >> Y;
  LL ans;
  LL n, m;
  
  if ((2 * X - Y) % 3 != 0 || (2 * X - Y) < 0 || (2 * Y - X) % 3 != 0 || (2 * Y - X) < 0) {
    ans = 0;
  } else {
    n = (2 * X - Y) / 3;
    m = (2 * Y - X) / 3;
  }
  
  ans = com(n, m);
  cout << ans << endl;
  
  
}