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;
using int64 = long long;
#define int int64
#define rep(i, n) for(int i=0; i<n; i++)
#define FOR(i, a, b) for(int i=a; i<b; i++)
#define SORT(x) sort(x.begin(), x.end())
#define GSORT(x) sort(x.begin(), x.end(), greater<int>())
#define mk make_pair
#define fi first
#define se second
#define pb push_back
#define ALL(x) x.begin(), x.end()
#define V(T) vector<T>
typedef pair<int, int> P;
typedef pair<P, P> PP;
typedef vector<int> vi;
typedef vector<vi> vvi;
int max(int a, int b) {if(b>a) return b; else return a;}
int min(int a, int b) {if(b<a) return b; else return a;}

const int MOD = 1000000007;
const int MAX = 700050;
int fac[MAX], finv[MAX];

int mmul(int a, int b) {
    return ((a%MOD)*(b%MOD))%MOD;
}

int mpow(int a, int b) {
    int res = 1;
    while(b) {
        if(b & 1) res = (res*a) % MOD;
        b >>= 1;
        a = (a*a) % MOD;
    }
    return res;
}

int mdiv(int a, int b) {
    return (a * mpow(b, MOD-2)) % MOD;
}

int mcomb(int n, int k) {
    if(n<k || k<0) return 0;
    else return mmul(mmul(fac[n], finv[k]), finv[n-k]);
}

inline void INIT_F(){
    fac[0] = finv[0] = 1;
    for(int i=1; i<MAX-1; i++) {
        fac[i] = (fac[i-1]*i) % MOD;
        finv[i] = mdiv(finv[i-1], i);
    }
}

signed main() {
    int X, Y;
    cin >> X >> Y;
    INIT_F();

    if((X+Y)%3) cout << 0 << endl;
    else {
        int n, m;
        n = (2*X-Y)/3;
        m = (2*Y-X)/3;
        if(n<0 || m<0) {
            cout << 0 << endl;
            return 0;
        }

        int res = mcomb(n+m, n);
        cout << res << endl;
    }

    
    return 0;
}