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 <iostream>
#include <string>
#include <vector>
#include <algorithm>
#include <stack>
#include <queue>
#include <cmath>
#include <tuple>
#include <cstdio>
#include <bitset>
#include <sstream>
#include <iterator>
#include <numeric>
#include <map>
#include <cstring>
#include <set>
#include <functional>
#include <iomanip>

using namespace std;

#define DEBUG_ //!!提出時にコメントアウト!!
#ifdef DEBUG_
	#define dump(x)  cerr << #x << " = " << (x) << endl;
#else
	#define dump(x)  ;
#endif

#define equals(a,b) (fabs((a)-(b)) < EPS)
#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define SZ(x) ((int)(x).size())
#define pb push_back
#define eb emplace_back

//#define int long long

typedef long long LL;
typedef vector<int> VI;
typedef vector<LL> VL;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;

template <typename T>
std::string printVector(const std::vector<T> &data)
{
    std::stringstream ss;
    std::ostream_iterator<T> out_it(ss, ", ");
    ss << "[";
    std::copy(data.begin(), data.end() - 1, out_it);
    ss << data.back() << "]";
    return ss.str();
}

template <typename T>
void print_array(const T &ary, int size){
    REP(i,size){
        cout << ary[i] << " ";
    }
    cout << endl;
}

const LL LINF = 1001002003004005006ll;
const int INF = 1001001001;
const double EPS = (1e-10);

const int MAX = 1e6+10;
const int MOD = 1000000007;

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

// テーブルを作る前処理
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(int argc, char const *argv[])
{
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout << setprecision(12);
    COMinit();

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

    int h = (X+Y) / 3;

    if(abs(X-Y) > h){
        cout << 0 << endl;
        return 0;
    }

    int n = abs(abs(X-Y) - h) / 2;
    cout << COM(h,n) << endl;


}