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 9.2.1) (judge verdict: Accepted)

/**
*	created: 21.08.2020 15:37:56
**/
#include <bits/stdc++.h>
// #include <boost/multiprecision/cpp_int.hpp>
// using bint = boost::multiprecision::cpp_int;
using namespace std;
using ll = long long;
using P = pair<int,int>;
// #define endl '\n'
#define int long long
#define rep(i,n) for (int i = 0; i < (int)(n); i++)
#define rep2(i,s,n) for (int i = (s); i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define pcnt(bit) __builtin_popcountll(bit)
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;}
const long double pi = acos(-1.0);
const int MAX = 1000010;
const int INF = LLONG_MAX;
const int MOD = 1000000007;
// const int MOD = 998244353;

template <typename T> T pow(T a, long long n, T e = 1) {
    T ret = e;
    while (n) {if (n & 1) ret *= a; a *= a; n >>= 1;}
    return ret;
}

template <int mod> struct ModInt {
    int x;
    ModInt() : x(0) {}
    ModInt(long long x_) {if ((x = x_ % mod + mod) >= mod) x -= mod;}
    ModInt& operator+=(ModInt rhs) {if ((x += rhs.x) >= mod) x -= mod; return *this;}
    ModInt& operator-=(ModInt rhs) {if ((x -= rhs.x) < 0) x += mod; return *this;}
    ModInt& operator*=(ModInt rhs) {x = (unsigned long long)x * rhs.x % mod; return *this;}
    ModInt& operator/=(ModInt rhs) {x = (unsigned long long)x * rhs.inv().x % mod; return *this;}
    ModInt operator-() const {return -x < 0 ? mod - x : -x;}
    ModInt operator+(ModInt rhs) const {return ModInt(*this) += rhs;}
    ModInt operator-(ModInt rhs) const {return ModInt(*this) -= rhs;}
    ModInt operator*(ModInt rhs) const {return ModInt(*this) *= rhs;}
    ModInt operator/(ModInt rhs) const {return ModInt(*this) /= rhs;}
    bool operator==(ModInt rhs) const {return x == rhs.x;}
    bool operator!=(ModInt rhs) const {return x != rhs.x;}
    ModInt inv() const {return pow(*this, mod - 2);}
    friend ostream& operator<<(ostream& s, ModInt<mod> a) {s << a.x; return s;}
    friend istream& operator>>(istream& s, ModInt<mod>& a) {s >> a.x; return s;}
};

using mint = ModInt<MOD>;

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

void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    rep2(i,2,MAX){
        fac[i] = fac[i-1] * (mint)i;
        inv[i] = (mint)MOD - inv[MOD%i] * (mint)(MOD/i);
        finv[i] = finv[i-1] * inv[i];
    }
}

mint COM(int n,int k) {
    if (n < k or n < 0 or k < 0) return 0;
    return fac[n] * finv[k] * finv[n-k];
}

signed main() {

    cin.tie(0);
    ios_base::sync_with_stdio(false);

    int a, b; cin >> a >> b;
    if ((a + b) % 3 != 0) {cout << 0 << endl; return 0;}
    int n = (a + b) / 3;
    COMinit();
    mint ans = COM(n,a-n);
    cout << ans << endl;
    return 0;
}