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 <set>
#include <algorithm>
#include <cctype>
#include <cmath>
#include <queue>
#include <map>
#include <numeric>
#include <unordered_map>
#include <iomanip>
#include <functional>
#include <bitset>
#include <complex>
#include <stack>



#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define rrep(i, n) for(int i = (int)(n-1); i >= 0; i--)
#define repi(i,a,b) for(int i=int(a);i<int(b);i++)
#define all(x) (x).begin(),(x).end()

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; }

typedef long long ll;

using namespace std;
long long modpow(long long x,long long n,long long mod)
{
  if(n==0)return 1;
  long long res=modpow(x*x%mod,n/2,mod);
  if(n&1)res=res*x%mod;
  return res;
}
long long modinv(long long a,long long mod)
{
  return modpow(a,mod-2,mod);
}
vector<long long> fact,inv;
long long mod=1e9+7;
void init(int n){
  fact.resize(n);
  inv.resize(n);
  fact[0]=1LL;
  rep(i,n-1){
    fact[i+1]=(long long)fact[i]*(i+1)%mod;
  }
  inv[n-1]=modinv(fact[n-1],mod);
  rrep(i,n-1){
    inv[i]=(long long)inv[i+1]*(i+1)%mod;
  }
}
long long comb(ll a,ll b){
  return fact[a]*inv[b]%mod*inv[a-b]%mod;
}

int main()
{
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll x,y;cin>>x>>y;
  if((x+y)%3!=0){
    cout<<0<<endl;
    return 0;
  }
  ll a,b;
  a=y-(x+y)/3;
  b=x-(x+y)/3;
//  cout<<a<<" "<<b<<endl;
  if(a<0||b<0){
    cout<<0<<endl;
    return 0;
  }
  init(2e6);
  cout<<comb(a+b,a)<<endl;


  return 0;

}