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>
#include<stdlib.h>
#include <math.h>
#define M 1000000007
#define ll long long

ll fact[3000001],ifact[3000001];

ll mpow(ll x,ll n){
  ll ans = 1;
  while(n>0){
    if(n&1) ans=ans*x%M;
    x = x*x%M;
    n=n>>1;
  }
  return ans;
}

ll combi(ll n,ll r){
  return fact[n]*ifact[r]%M*ifact[n-r]%M;
}

int main(){
  ll x,y,k,m,n;
  scanf("%lld %lld",&x,&y);
  if((2*x-y)%3!=0||(2*y-x)%3!=0){
  	printf("0");
    return 0;
  }
  n=(2*x-y)/3;
  m=(2*y-x)/3;
  fact[0]=1;
  ifact[0]=1;

  for(ll i=1;i<=m+n;i++){
    fact[i]=fact[i-1]*i%M;
    ifact[i]=ifact[i-1]*mpow(i,M-2)%M;
  }

  printf("%lld\n",combi(n+m,n));
  return 0;
}