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: Accepted)

#include <bits/stdc++.h>
#define r(i,n) for(long long i = 0; i<n; i++)
typedef long long ll;
using namespace std;

ll gyaku(ll a){
  ll d=1,e=1;
  r(i,15){
    d*=a;
    d%=1000000007;
  }
  r(i,66666667){
    e*=d;
    e%=1000000007;
  }
  return e;
}

int main() {
  ll x,y,a,b;
  cin >> x >> y;
  a=(2*x-y)/3;
  b=(2*y-x)/3;
  if(a+2*b!=y||b+2*a!=x||a<0||b<0){
   cout<<0<<endl;
  }
  else { ll A=1,B=1,C=1;
    a=(2*x-y)/3;
    b=(2*y-x)/3;
    r(i,a+b){
      C*=(i+1);
      C%=1000000007;
      if(i==a-1)A=C;
      if(i==b-1)B=C;
    }
    ll f = (C*gyaku(A)%1000000007)*gyaku(B)%1000000007;
    cout<<f<<endl;
       }
}