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: Accepted)

/*
　　　∫ ∫ ∫
　　　ノヽ
　　（＿　 ）
　（＿　　　 ）
（＿＿＿＿＿＿ ）
　ヽ(´･ω･)ﾉ　
　　 |　 /
　　 UU
*/
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
using P = pair<ll, ll>;
typedef vector<int> vi;
const ll INF = 1LL << 60;
const int inf = 1<<30;
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
#define REP(i, n) for (int i = 0; i < (n); i++)
#define FOR(i,s,n) for (int i = s; i < (n); i++)
#define ALL(obj) (obj).begin(), (obj).end() //コンテナじゃないと使えない!!
#define debug(x) cerr << #x << ": " << x << "\n";
#define mp make_pair
template <typename T>
ostream& operator<<(ostream& os, vector<T> &V){
    int N = V.size();
    REP(i,N){
        os << V[i];
        if (i!=N-1) os << " ";
    }
    os << "\n";
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, pair<T,T> const&P){
    os << P.first;
    os << " ";
    os << P.second;
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, set<T> &S){
    auto it=S.begin();
    while(it!=S.end()){
        os << *it;
        os << " ";
        it++;
    }
    os << "\n";
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, deque<T> &q){
    for(auto it=q.begin();it<q.end();it++){
        os<<*it;
        os<<" ";
    }
     os<<endl;
    return os;
}

const int MAX = 5000000;
const int MOD = 1000000007;

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}

// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
//fixed<<setprecision(10)<<ans<<endl;
int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    ll X,Y;
    cin >> X >> Y;
    if ((X+Y)%3 != 0){
        cout << 0 << endl;
        return 0;
    }
    ll S = (X+Y);
    X -= S/3;
    Y -= S/3;
    COMinit();
    ll ans = COM(X+Y,X);
    cout << ans << endl;
}