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)

/*Function Template*/
#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define rep(i, n) for(ll i = 0; i < (n); i++)
#define ALL(a)  (a).begin(),(a).end()
#define IOS ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);

const int MAX = 510000;
const int MOD = 1000000007;

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

void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (ll 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;
    }
}

ll COM(ll n, ll k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}







/*↓↓↓↓↓↓↓↓↓↓↓↓↓↓↓*/
int main() {
  IOS;
  COMinit();
  ll x,y;
  cin>>x>>y;
  if((x+y)%3!=0 || 2*x<y || 2*y<x){
    cout<<0<<endl;
    return 0;
  }
  ll big,small;
  if(x>=y){
    big=x;
    small=y;
  }else{
    big=y;
    small=x;
  }
  ll n=0,k=0;
  ll bigg=big,smalll=small;
  rep(i,bigg){
    big-=1;
    small-=2;
    k+=1;
    if(big==2*small){
      n=small;
      break;
    }
  }
  n+=k;

  ll ans=1;
  for(ll i=1;i<=k;i++){
    ans*=n-(i-1);
    ans/=i;
    ans%=MOD;
  }
  cout<<COM(n,k)<<endl;
}