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: Accepted)

#include<stdio.h>
int mbk(long long a,int n,int m){return n?(mbk((a*a)%m,n/2,m)*(n%2?a:1))%m:1;}
int inv(int a,int m){return mbk((long long)a,m-2,m);}
int aCb(int a,int b,int m){
  long long s=1,i;
  if(b>a-b)b=a-b;
  for(i=0;i<b;i++)s=(((s*(a-i))%m)*inv(i+1,m))%m;  
  return (int)s;
}
int main(){
  long long n,m,s,M=1e9+7;
  scanf("%lld %lld",&n,&m);
  s=n+m;
  if(s%3)goto END;
  n-=s/3;
  m-=s/3;
  if(n<0||m<0)goto END;
  printf("%d\n",aCb(n+m,m,M));
  return 0;
 END:;
  printf("0\n");
  return 0;
}