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)

#include <bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define all(a) a.begin(), a.end()
#define pb push_back
#define mp make_pair
typedef long long ll;
typedef pair<ll,ll> P;
typedef pair<ll,int> pli;
typedef pair<int,int> pii;
#define rep(i,a,b) for(ll i=a ; i<b ; i++)
#define qrep(que, ite) for(auto ite=begin(que) ; ite!=end(que) ; ite++)
const int max_n = 1e5;
const ll mod = 1e9+7;
const ll INF = 1LL<<60;
const int inf = 1e5;
//typedef long double ld;
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; }
int dx[4] = {0,0,1,-1};
int dy[4] = {1,-1,0,0};
ll mo = 998244353;
struct UnionFind {
    vector<int> par;

    UnionFind(int n) : par(n, -1) { }
    void init(int n) { par.assign(n, -1); }

    int root(int x) {
        if (par[x] < 0) return x;
        else return par[x] = root(par[x]);
    }

    bool issame(int x, int y) {
        return root(x) == root(y);
    }

    bool merge(int x, int y) {
        x = root(x); y = root(y);
        if (x == y) return false;
        if (par[x] > par[y]) swap(x, y); // merge technique
        par[x] += par[y];
        par[y] = x;
        return true;
    }

    int size(int x) {
        return -par[root(x)];
    }
};

ll modpow(ll a, ll n, ll mod) {
    ll res = 1;
    while (n > 0) {
        if (n & 1) res = res * a % mod;
        a = a * a % mod;
        n >>= 1;
    }
    return res;
}
ll modinv(ll a, ll mod) {
    return modpow(a, mod - 2, mod);
}
const int MAX = 3000000;
const int MOD = 1000000007;
ll fac[MAX], finv[MAX], inv[MAX];
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;
    }
}
ll 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;
}
int main(){
  ll x,y; cin>>x>>y;
  if((2*x-y)%3 != 0 || (2*y-x)%3!=0){
    cout << 0 << endl;
  }else{
    ll a = (2*x-y)/3;
    ll b = (2*y-x)/3;
    COMinit();
    cout << COM(a+b, a) << endl;
  }
  return 0;
}