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>
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
//----------------------------------------------------------------------

using namespace std;
using ll = long long;
using P = pair<int,int>;
using Graph = vector<vector<int>>;

const int mod = 1'000'000'007;
struct mint {
    ll x;
    mint(ll x = 0) : x ((x%mod + mod)  %mod) {}
    mint operator - () {return mint(-x);}
    mint & operator +=(const mint a) {
        (x += a.x) %= mod;
        return *this;
    }
    mint & operator -=(const mint a) {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint & operator *=(const mint a) {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator + (const mint a) const {return mint(*this)+=a;}
    mint operator - (const mint a) const {return mint(*this)-=a;}
    mint operator * (const mint a) const {return mint(*this)*=a;}

    mint pow(ll t) const {
        mint res(1);
        mint a(x);
        while(t > 0) {
            if(t&1) {
                res *= a;
            }
            a *= a;
            t >>= 1;
        }
        return res;
    }

    mint inv() const {return pow(mod-2);}
    mint & operator /=(const mint a) {
        *this *= a.inv();
        return *this;
    }
    mint operator / (const mint a) { return mint(*this) /= a;}
};

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

int main(void) {
    int x, y;
    cin >> x >> y;
    if((x +y) % 3 != 0) {
        cout << 0 << endl;
        return 0;
    }

    int n = (x + y) / 3;
    int u = y - n;

    mint ans = choose(n, u);

    cout << ans.x << endl;
    return 0;
}