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: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>

typedef long long ll;

ll modnum=7+1000000000;

ll modpow(ll a, ll n, ll mod){
    if(n==0) return 1;
    if(n%2==0){
        return modpow(a*a%mod, n/2, mod);
    }else{
        return a*modpow(a, n-1, mod)%mod;
    }
}

int main(void){
    ll X, Y;
    ll a, b;
    ll ans=0;
    ll XpY=1;
    ll Xa=1;
    ll Ya=1;
    ll Xinv;
    ll Yinv;
    

    scanf("%lld %lld", &X, &Y);

    if((2*X-Y)%3!=0 || (2*Y-X)%3!=0/* || (2*X-Y)<=0 || (2*Y-X)<=0*/){
        printf("0");
        return 0;
    }
    a=(2*X-Y)/3;
    b=(2*Y-X)/3;


    ll i;
    for(i=1; i <= a; i++){
        Xa=(Xa*i)%modnum;
    }
    for(i=1; i <= b; i++){
        Ya=(Ya*i)%modnum;
    }
    for(i=1; i <= a+b; i++){
        XpY=(XpY*i)%modnum;
    }
    
    Xinv=modpow(Xa, modnum-2, modnum);
    Yinv=modpow(Ya, modnum-2, modnum);

    ans = (XpY*Xinv)%modnum;
    ans = (ans*Yinv)%modnum;

    printf("%lld", ans%modnum);
    return 0;
}