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)

#pragma GCC optimize("O3")
#include <iostream>
#include <iomanip>
#include <cstdio>
#include <string>
#include <cstring>
#include <deque>
#include <list>
#include <queue>
#include <stack>
#include <vector>
#include <utility>
#include <algorithm>
#include <map>
#include <set>
#include <complex>
#include <cmath>
#include <limits>
#include <cfloat>
#include <climits>
#include <ctime>
#include <cassert>
#include <numeric>
#include <fstream>
#include <functional>
#include <bitset>
using namespace std;

using ll = long long;
using P = pair<int, int>;
using T = tuple<int, int, int>;

template <class T> inline T chmax(T &a, const T b) {return a = (a < b) ? b : a;}
template <class T> inline T chmin(T &a, const T b) {return a = (a > b) ? b : a;}

constexpr int MOD = 1e9 + 7;
constexpr int inf = 1e9;
constexpr long long INF = 1e18;
constexpr double pi = acos(-1);
constexpr double EPS = 1e-10;

int dx[] = {1, 0, -1, 0};
int dy[] = {0, 1, 0, -1};

ll modpow(ll a, ll b){
    if(b == 0) return 1;
    else if(b % 2 == 0){
        ll d = modpow(a, b/2) % MOD;
        return (d * d) % MOD;
    }
    else{
        return (a * modpow(a, b-1)) % MOD;
    }
}

const int MAX_N = 1000011;

ll fact[MAX_N], finv[MAX_N];
ll comb(int n, int r){
    if(n < r || r < 0) return 0;
    return fact[n] * finv[n-r] % MOD * finv[r] % MOD;
}

void calc(int n){
    fact[0] = finv[0] = 1;
    for(int i=1; i<=n; i++){
        fact[i] = (fact[i-1] * i) % MOD;
        finv[i] = modpow(fact[i], MOD-2);
    }
}

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

    calc(1000010);

    int x, y; cin>>x>>y;

    int minnum = -1, maxnum = -1;
    for(int i=0; i<=1000000; i++){
        int X = x, Y = y;
        X -= i;
        Y -= i * 2;

        if(0 <= X && 0 <= Y && Y * 2 == X){
            minnum = i;
            maxnum = i + Y;
            break;
        }
    }

    //cout << maxnum << endl;

    cout << comb(maxnum, minnum) << endl;

    return 0;
}