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 rep(i, n) for (int i = 0; i < n; ++i)
#define stl_rep(itr, x) for (auto itr = x.begin(); itr != x.end(); ++itr)
#define stl_repr(itr, x) for (auto itr = x.rbegin(); itr != x.rend(); ++itr)
#define all(x) x.begin(), x.end()
#define allr(x) x.rbegin(), x.rend()

using llong = long long int;

const static int MOD = 1000000007;
const static int IINF = INT_MAX;
const static llong LINF = LLONG_MAX;
const static int dx[4] = {1, 0, -1, 0};
const static int dy[4] = {0, 1, 0, -1};

template<typename T> T my_gcd(T a, T b) {
    return (a ? my_gcd(b % a, a) : b);
}
template<typename T> T my_lcm(T a, T b){
    return a * b / my_gcd(a, b);    
}

const int MAX = 9999999;

llong 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;
    }
}

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;
}

int main (int argc, char *argv[]) {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int x, y;
    cin >> x >> y;
    if ((2 * y - x) % 3 != 0 || (2 * x - y) % 3 != 0) {
    	cout << 0 << endl;
    	return 0;
    }

    COMinit();
    int n = (2 * y - x) / 3, m = (2 * x - y) / 3;
    cout << COM(n + m, n) << endl;

    return 0;
}