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 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<stdio.h>
int main(void){
  int X,Y,x,y,i,j,c,n,k;
  scanf("%d %d",&X,&Y);
  if ((2*X-Y)%3==0){
  y=(2*X-Y)/3;
  x=X-2*y;
  printf("%d %d\n",x,y);
  n=1;
  k=1;
  for(i=x+y;i>y;i--){
    n=n*i;
  }
  /*  printf("%d\n",n); */
  for(i=x;i>0;i--){
    k=k*i;
  }
  /*printf("%d\n",k);*/
    c=(int)n/k;
  printf("%d\n",c);
  }
  if((2*X-Y)%3!=0){
    printf("0\n");
  }
    return 0;
}