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: Runtime Error)

#include<bits/stdc++.h>
using namespace std;

#pragma GCC optimze("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,avx2,tune=native")
#pragma GCC optimize("fast-math")
#pragma GCC optimize("unroll-loops")

#define x first
#define y second
#define pb push_back
#define mp make_pair

#define all(a) (a).begin(), (a).end()
#define sz(a) (int) (a).size()

#define ll long long
#define int long long
#define ld long double

const int N = 4e6 + 10;

const int M = 1e9 + 7;

int fact[N];

int pw(int a, int n) {
    int res = 1;
    while (n>0) {
        if (n & 1) {
            res *= a;
            res %= M;
            n--;
        }else {
            a *= a;
            a %= M;
            n >>= 1;
        }
    }
    return res;
}

int inv(int a) {
    return pw(a, M - 2);
}

int32_t main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int x, y;
    cin >> x >> y;
    int z = x + y;
    if (z % 3 != 0) {
        cout << "0\n";
        return 0;
    }
    int c = z / 3;
    x -= c;
    y -= c;
    //cout << x << " " << y << '\n';
    fact[0] = 1;
    for (int i = 1; i < N; i++) fact[i] = (fact[i - 1] * i) % M;
    cout << (((((fact[c] * inv(fact[x + y])) % M * inv(fact[c - x - y])) % M * fact[x + y]) % M * inv(fact[x])) % M * inv(fact[y])) % M << '\n';
    return 0;
}