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: Wrong Answer)

#pragma GCC optimize("Ofast")   //Comment optimisations for an interatcive problem - use endl
#pragma GCC target("avx,avx2,fma")
#pragma GCC optimization ("unroll-loops")
#include<bits/stdc++.h>

using namespace std;

#define  fastio         ios:: sync_with_stdio(0);cin.tie(0);cout.tie(0);cout<<fixed;cout<<setprecision(10);
#define  randomINIT     mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());
#define  foo(i,a,n)     for(ll i = (a); i <= n; i++)
#define  frr(i,a,n)     for(ll i = (a); i >= n; i--)
#define  fo(i,n)        for(ll i = 0; i < n; i++)
#define  all(x)         (x).begin(),(x).end()
#define  mset(x,val)    memset(x,val,sizeof(x))
#define  newl           cout<<"\n"
#define  pb             push_back
#define  mp             make_pair
#define  s              second
#define  f              first
#define  dline          cout<<"///REACHED///\n";
#define  deb1(x)        cout<<#x<<" = "<<x<<'\n';
#define  deb2(x,y)      cout<<'['<<#x<<','<<#y<<"] = "<<'['<<x<<','<<y<<']'<<'\n';
#define  deb3(x,y,z)    cout<<'['<<#x<<','<<#y<<','<<#z<<"] = "<<'['<<x<<','<<y<<','<<z<<']'<<'\n';

typedef  long long             ll;
typedef  long double           ld;
typedef  pair<ll,ll>           pll;
typedef  vector<ll>            vll;
typedef  unordered_map<ll,ll>  um;
typedef  vector<pll>           vpll;

const ll   MOD   =  1e+9+7;
const ll   INF   =  0x7f7f7f7f7f7f7f7f;
const int  INFi  =  0x7f7f7f7f;
const ll   MAXN  =  1e+6+7;

vll adj[MAXN];ll visit[MAXN]={};
int dx8[]={0,1,1,1,0,-1,-1,-1}, dy8[]={1,1,0,-1,-1,-1,0,1};
int dx4[]={0,1,0,-1}, dy4[]={1,0,-1,0};

//<<-----Declare Variable Here------->>//
int qq;
ll x, y;
ll n, r;
ll ff[MAXN];
//<<-----Implement Functions Here---->>//
ll mult(ll a, ll b) {return (a * b)%MOD;}

ll fastModInv(ll x){
	ll res=1;
	ll y=MOD-2;
	while(y){
		if(y&1)res=mult(res,x);
		y>>=1;x=mult(x,x);
	}
	return res;
}

ll nCr(){
	return mult(fastModInv(ff[r]),mult(ff[n],fastModInv(ff[n-r])));
}
//<<-----Start of Main--------------->>//

int main(){
	fastio;randomINIT;
	ff[0]=0;ff[1]=1;
	foo(i,2,MAXN-2)ff[i]=mult(ff[i-1],i);

	cin>>x>>y;	
	if((x+y)%3!=0){cout<<0;return 0;}
	n=(x+y)/3;r=x-n;
	//cout<<n<<" "<<r;newl;
	//now find nCr
	if(y>2*n || y<n || x<n || x>2*n){cout<<0;return 0;}
	cout<<nCr();


    #ifndef ONLINE_JUDGE
    cout<<"\nTime Elapsed: " << 1.0*clock() / CLOCKS_PER_SEC << " sec\n";
    #endif
}