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<iostream>
typedef long long ll;
using namespace std;

const int MOD = 1e9 + 7, MAX = 2500000;

int comb(int n, int k){
    // Calculate factorial and inverse element
    ll fac[MAX], finv[MAX], inv[MAX];
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }

    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int main() {
    int X, Y, ans = 0;
    cin >> X >> Y;
    int n = (2 * Y - X) / 3;
    int m = (2 * X - Y) / 3;
    if ((X + Y) % 3 == 0 && n >= 0 && m >= 0) {
        ans = comb(n + m, n);
    }
    cout << ans << endl;
}