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)

#include <bits/stdc++.h>
using namespace std;
#define ll int64_t
#include <cstdlib> 
#include <math.h>
#include<cstdio>
#include<cstring>
#define dbg(x) cout << #x << " = " << x << endl
#define dbg2(x,y) cout << #x << " = " << x << ", " << #y << " = " << y << endl
#define dbg3(x,y,z) cout << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << endl
#define dbg4(x,y,z,q) cout << #x << " = " << x << ", " << #y << " = " << y << ", " << #z << " = " << z << ", " << #q << " = " << q << endl

#define FOR(I, A, B) for (int I = (A); I <= (B); I++)
typedef vector< int > vi;
typedef vector< vi > vvi;
typedef pair< int,int > ii;
#define gap cout<<endl;
#define IOS ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
#define fo(i,n) for(int i=0;i<n;i++)
#define sz(a) ll((a).size())
#define pb push_back
#define all(c) (c).begin(),(c).end()
#define tr(c,i) for(typeof((c)).begin() i = (c).begin(); i != (c).end(); i++)
#define present(c,x) ((c).find(x) != (c).end())
#define cpresent(c,x) (find(all(c),x) != (c).end())
const ll N=1000005;
ll fact[N];

const ll MOD=1000000007;
ll pow(ll a, ll b, ll m)
{
    ll ans=1;
    while(b)
    {
        if(b&1)
            ans=(ans*a)%m;
        b/=2;
        a=(a*a)%m;
    }
    return ans;
}

ll modinv(ll k)
{
    ll ans=pow(k, MOD-2, MOD);
    ans=(ans+MOD)%MOD;
    return ans;
}

void precompute()
{
    fact[0]=fact[1]=1;
    for(ll i=2;i<N;i++)
    {
        fact[i]=fact[i-1]*i;
        fact[i]%=MOD;
    }
    
}

ll nCr(ll x, ll y)
{
    
    if(y>x)
        return 0;
    ll num=fact[x];
    num*=modinv(fact[y]);
    num%=MOD;
    num*=modinv(fact[x-y]);
    num%=MOD;
    return (num+MOD)%MOD;
}


int main(){
    precompute();
    ll x,y;
    cin>>x>>y;
    ll n,m;
    if((x+y)%3!=0){cout<<0<<endl; return 0;}
    if(x<0||y<0){
        cout<<0<<endl;
        return 0;
    }
    m=(2*x-y)/3;
    n=(2*y-x)/3;
    m=m%MOD;
    n=n%MOD;
    
    ll ans=nCr((n+m)%MOD,n);
    ans=(ans+MOD)%MOD;
    //dbg(invfact[1]);
    cout<<ans<<endl;
    
    
}