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# (Mono 4.6.2.0) (judge verdict: Wrong Answer)

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Threading.Tasks;

namespace CSharp_practice
{
    class Program
    {
        static void Main(string[] args)
        {
            int max = 700000;
            int mod = 1000000007;
            long[] fac = new long[max];
            long[] finv = new long[max];
            long[] inv = new long[max];

            long[] input = Console.ReadLine().Split(' ').Select(long.Parse).ToArray();
            long x = input[0];
            long y = input[1];

            if (x + y % 3 != 0)
            {
                Console.WriteLine(0);
            }
            else
            {
                long m = (2 * x - y) / 3;
                long n = (2 * y - x) / 3;
                COMinit(max, mod, fac, finv, inv);
                Console.WriteLine(COM(mod, fac, finv, inv, m + n, n));
            }
        }

        static void COMinit(int max, int mod, long[] fac, long[] finv, long[] inv)
        {
            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;
            }
        }

        static long COM(int mod, long[] fac, long[] finv, long[] inv, long n, long k)
        {
            if (n < k) return 0;
            if (n < 0 || k < 0) return 0;
            return fac[n] * (finv[k] * finv[n - k] % mod) % mod;
        }
    }
}