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++ (GCC 9.2.1) (judge verdict: Runtime Error)

//----------BHAVIK DIWANI(PICT_COMP)---------------
#include<bits/stdc++.h>
#include<stdlib.h>
#include<stdio.h>
#define test ll t; cin>>t; while(t--)
#define FOR(p,n) for(ll i=p;i<(ll)n;i++)
#define fastio ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);
#define mod 1000000007
#define ll long long
#define int long long
#define ull unsigned long long
#define MAX 1000005
#define pb push_back
#define mkp make_pair
#define vi vector<int>
#define pii pair<int,int>
#define endl '\n'
#define vs vector<string>
#define mii map<int,int>
#define msi map<string,int>
#define vpii vector< pair<int, int > >
#define vpsi vector< pair< string ,int > >
#define forci(p,q) for(int i=p;i<q;i++)
ll gcd(ll a,ll b){ if(b==0) return a;    return gcd(b,a%b);}
ll lcm(ll a,ll b) { return (a/gcd(a,b)*b);}
ull power(ull a, ull b) {a %= mod;  ull res = 1;  while (b > 0) { if (b & 1) res = res * a % mod;  a = a * a % mod;  b >>= 1; }  return res%mod; }
ll modmul(ll x, ll y){return (x*y)%mod;}
ll modadd(ll x, ll y){return (x+y)%mod;}
ll modsub(ll x, ll y){return (x-y+mod)%mod;}
ll fact[1000009]={0};
void facto() {fact[0]=1;fact[1]=1;for(int i=2;i<=1000008;i++)fact[i]=(fact[i-1]*i)%mod;}
ll ncr(ll n,ll r) {ll res=1; res=fact[n]; res=(res*(power(fact[r],mod-2)))%mod; res=(res*(power(fact[n-r],mod-2)))%mod; return res;}
ll npr(ll n,ll r) {ll res=1; res=fact[n]; res=(res*(power(fact[n-r],mod-2)))%mod; return res;}
inline long long toint(const std::string &s) {std::stringstream ss; ss << s; long long x; ss >> x; return x;}
inline std::string tostring(long long number) {std::stringstream ss; ss << number; return ss.str();}
inline std::string tobin(long long x) {return std::bitset<63>(x).to_string();}
/*bool prime[MAX];
ull sieve(){memset(prime,true,sizeof(prime));for(ull p=2;p*p<MAX;p++){if(prime[p]==true){for(ull i=2*p;i<MAX;i+=p){prime[i]=false;}}}prime[0]=0;prime[1]=0;
}*/
/*int inv[1000001]={0};
void findinverse(){inv[1] = 1;for(int i = 2; i <=1e6 ; ++i)	inv[i] = (mod - (mod/i) * inv[mod%i] % mod) % mod;}
*/
using namespace std;
int solve()
{
	facto();
	int x,y;
	cin>>x>>y;
	if((2*x-y)%3==0 && (2*y-x)%3==0)
	{
		int moves=(x+y)/3;
		int n=2*moves-x;
		int m=moves-n;
		int ans=ncr(moves,n)%mod;
		cout<<ans<<endl;
	}
	else
		cout<<"0";
	return 0;
}
signed main()
{
	fastio;
//	test
		solve();
}