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: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define pb push_back
const int maxn = 10000;
const int INF32 = 1'050'000'000;
const long long INF64 = 4'000'000'000'000'000'000;
const int MOD7 = 1'000'000'007;
const int MOD9 = 1'000'000'009;
const int mod = MOD7;

void ERROR(int num) { cout << "ERROR" << num << endl; }
ll gcd(ll a, ll b) { return b ? gcd(b, a%b) : a; }
int dx[8]={-1,0,1,0,1,1,-1,-1};
int dy[8]={0,-1,0,1,1,-1,1,-1};

struct mint{
    using ull=unsigned long long int;
    ull v;
    mint(ll vv=0){s(vv%mod+mod);}
    mint& s(ull vv){
        v=vv<mod?vv:vv-mod;
        return *this;
    }
    //オーバーロード
    mint operator-()const{return mint()-*this;}//符号反転
    mint&operator+=(const mint&val){return s(v+val.v);}
    mint&operator-=(const mint&val){return s(v+mod-val.v);}
    mint&operator*=(const mint&val){
        v=ull(v)*val.v%mod;
        return *this;
    }
    mint&operator/=(const mint&val){return *this*=val.inv();}
    mint operator+(const mint&val){return mint(*this)+=val;}
    mint operator-(const mint&val){return mint(*this)-=val;}
    mint operator*(const mint&val){return mint(*this)*=val;}
    mint operator/(const mint&val){return mint(*this)/=val;}
    mint pow(int n)const{
        mint res(1),x(*this);
        while(n){
            if(n&1)res*=x;
            x*=x;
            n>>=1;
        }
        return res;
    }
    mint inv()const{return pow(mod-2);}
    //拡張ユークリッドの互除法
    /* mint inv()const{
        int x,y;
        int g=extgcd(v,mod,x,y);
        assert(g==1);
        if(x<0)x+=mod;
        return mint(x);
    }*/
    friend ostream& operator<<(ostream&os,const mint&val){
        return os<<val.v;
    }//出力
    bool operator<(const mint&val)const{return v<val.v;}
    bool operator==(const mint&val)const{return v==val.v;}
    bool operator>(const mint&val)const{return v>val.v;}
};

mint choose(ll n, ll a){
    mint x = 1, y = 1;
    rep(i,a){
        x *= n-i;
        y *= i+1;
    }
    return x / y;
}

int main(){
    std::ios::sync_with_stdio(false);
    std::cin.tie(0);

    ll X, Y;
    cin >> X >> Y;

    if((X+Y)%3==0){
      ll n = (X+Y)/3;
      X-=n, Y-=n;
      cout << choose(X+Y,X) << endl;
    }
    else{
      cout << 0 << endl;
    }
}