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 ll long long                                    // long long省略
#define pb push_back                                    // push_back省略
#define mp make_pair                                    // make_pair省略
#define fi first                                        // first省略
#define se second                                       // second省略
#define itn int                                         // int誤字保険
#define count cout                                      // cout誤字保険
#define vecotr vector                                   // vector誤字保険
#define ednl endl                                       // endl誤字保険
#define opt() cin.tie(0); ios::sync_with_stdio(false)   // 入出力速度改善

#define rep(i,l,r) for(ll i=(l);i<(r);i++)               // 範囲[l, r)で刻み1のfor文(順方向)
#define repp(i,l,r,k) for(ll i=(l);i<(r);i+=(k))         // 範囲[l, r)で刻みkのfor文(順方向)
#define rrep(i,l,r) for(ll i=(r-1);i>=(l);i--)           // 範囲[l, r)で刻み1のfor文(逆方向)
#define rrepp(i,l,r,k) for(ll i=(r-1);i>=(l);i-=(k))     // 範囲[l, r)で刻みkのfor文(逆方向)
#define all(x) (x).begin(), (x).end()                   // vectorのポインタ位置指定用
#define max(p,q)((p)>(q)?(p):(q))                       // max拡張
#define min(p,q)((p)<(q)?(p):(q))                       // min拡張
#define bit(n,m)(((n)>>(m))&1)                          // 変数nのm番目のbitを取り出す

template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }

int dy[]={0, 1, 0, -1};                                 // 4方向近傍
int dx[]={1, 0, -1, 0};                                 // 4方向近傍

const int MAX = 10000000;
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){
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

////////////////////////////////////////////////////////////////////////////////////////

int main(){
    ll x,y;cin>>x>>y;
    COMinit();
    ll tmp = (x+y)/3, cnt = -1;
    rep(i,0,tmp+1){
        if(i*2+(tmp-i)*1 == x && i*1+(tmp-i)*2 == y){
            cnt = i;
            break;
        }
    }
  	if(cnt == -1){
      cout << 0 << ednl;
      return 0;
    }
  	//cout << cnt << ednl;
    cout << COM((x+y)/3,cnt) % MOD << ednl;
}