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<bits/stdc++.h>
using namespace std;
#define rep(i,n) for(ll i=0;i<n;i++)
#define repl(i,l,r) for(ll i=(l);i<(r);i++)
#define per(i,n) for(ll i=n-1;i>=0;i--)
#define lper(i,r,l) for(ll i=r-1;i>=l;i--)
#define fi first
#define se second
#define mp make_pair
#define all(x) (x).begin(),(x).end()
#define CST(x) cout<<fixed<<setprecision(x)
using ll=long long;
using vl=vector<ll>;
using vvl=vector<vector<ll>>;
using pl=pair<ll,ll>;
const ll MOD=1000000007;
const ll MOD9=998244353;
const int inf=1e9+10;
const ll INF=4e18;
const ll dy[4]={1,0,-1,0};
const ll dx[4]={0,-1,0,1};
long long fac(ll a){
    long long ans=1;
    while(a>1){
        ans*=a;
        ans%=1000000007;
        a--;
    }
    return ans;
}
long long modinv(long long a, long long m) {
    long long b = m, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= m; 
    if (u < 0) u += m;
    return u;
}
int main(){
    ll a,b;
    cin >> a >> b;
    if((a+b)%3!=0){
        cout << 0 << endl;
        return 0;
    }
    ll c=(a+b)/3;
    if(2*c<b||2*b<a){
        cout << 0 << endl;
        return 0;
    }
    ll count=0;
    rep(i,c+1){
        if(2*c-i==a&&c+i==b)count=i;
    }
    count=min(count,c-count);
    ll ans=1;
    rep(i,count){
        ans*=c-i;
        ans%=MOD;
    }
    rep(i,count){
        ans*=modinv(i+1,MOD);
        ans%=MOD;
    }
    cout << ans << endl;
}