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;
#pragma GCC optimize("Ofast")
#define all(x) (x).begin(),(x).end()
#define rep(i,n) for(int (i)=0;i<(n);i++)
#define REP(i,m,n) for(int (i)=(m);(i)<(n);(i)++)
#define INF LLONG_MAX
#define MOD 1000000007
#define fcout cout << fixed << setprecision(15)
#define int long long
#define yorn(f) puts((f)?"Yes":"No")
#define YORN(f) puts((f)?"YES":"NO")
#define lower_bound lowb
#define upper_bound upb
typedef long long ll;
typedef pair<int, int> P;
typedef priority_queue<int> pque;
int gcd(int a,int b){return b?gcd(b,a%b):a;};
int lcm(int a,int b){return a*b/gcd(a,b);};
int mod(int a,int b){return (a+b-1)/b;};
int month[12] = {31,28,31,30,31,30,31,31,30,31,30,31};
template<typename A, size_t N, typename T>
void Fill(A(&array)[N],const T &val){std::fill((T*)array,(T*)(array+N),val);}
template<class T>inline bool chmax(T& a,T b){if(a<b){a=b;return true;}return false;};
template<class T>inline bool chmin(T& a,T b){if(a>b){a=b;return true;}return false;};

const int MAX = 1000010;
long long 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;
    }
}

// 二項係数計算
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;
}


signed main() {
	cin.tie(0);
	cout.tie(0);
	ios::sync_with_stdio(false);

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

	int m, n;
	n = ((2 * y - x) / 3) % MOD;
	m = ((2 * x - y) / 3) % MOD;

	if(m < 0 || n < 0) {
		cout << 0 << endl;
	} else {
		COMinit();
		cout << COM(n + m, n) << endl;
	}
	//cout << comb(x, y) << endl;
	return 0;
}