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;
 
const long long mod = 1e9 + 7ll ;
 
long long modpower(long long a , long long b)
{
    if(b == 0)
        return 1ll ;
    long long x = modpower(a , b / 2ll) ;
    x = (x * x) % mod ;
    if(b & 1)
        return (x * a) % mod ;
    return (x % mod) ;
}
 
struct combination
{
    vector<long long>fact , inv ;
    combination(int sz) : fact(sz + 1) , inv(sz + 1)
    {
        fact[0] = 1ll ;
        inv[0] = 1ll ;
        for(long long i = 1 ; i <= sz ; ++i)
        {
            fact[i] = (fact[i-1] * i) % mod ;
            inv[i] = modpower(fact[i] , mod - 2ll) ;
        }
    }
    long long choose(int n , int k) const
    {
        if(k < 0 || n < k)
            return 0 ;
        return ((((fact[n] * inv[k]) % mod) * inv[n - k]) % mod) ;
    }
};
 
combination comb(1e6+5) ;

int main()
{
    ios_base::sync_with_stdio(0) ;
    cin.tie(0) ;
    int x , y ;
    cin>>x>>y ;
    int curx = 0 , cury = 0 , moves = 0 ;
    long long ans = 0ll ;
    for(int i = 0 ; i <= x ; ++i)
    {
        curx = i ;
        cury = 2 * i ;
        if(curx > x || cury > y)
            break ;
        moves = y - cury ;
        curx += 2 * moves ;
        cury += moves ;
        if(curx != x || cury != y)
            continue ;
        ans = (ans + comb.choose(moves + i , i)) % mod ;
    }
    return cout<<ans<<"\n" , 0 ;
}