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 Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.util.Scanner;

public class Main {

	private static final int MAX = 1000001;
	private static final long[] fac = new long[MAX];
	private static final long[] finv = new long[MAX];
	private static final long[] inv = new long[MAX];
	private static final int MOD = 1_000_000_007;

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int x = sc.nextInt();
		int y = sc.nextInt();
		sc.close();

		int n = x+y;
		if(n%3 != 0){
			System.out.println(0);
			return;
		}
		COMinit();
		if(x == y * 2){
			System.out.println(COM(n/3,x-y));
		}else if(x == y){
			System.out.println(COM(n/3, n/6));
		}else if(x*2 == y) {
			System.out.println(COM(n/3,y-x));
		}else if(x > y){
			System.out.println(COM(n/3,x-y+1));
		}else{
			System.out.println(COM(n/3,y-x+1));
		}
	}

	private static void COMinit() {
		fac[0] = fac[1] = 1;
		finv[0] = finv[1] = 1;
		inv[0] = 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;
		}
	}
	private static long COM(int n,int k) {
		if(n < k) {
			return 0;
		}
		if(n < 0 || k < 0) {
			return 0;
		}
		return fac[n] * (finv[k] * finv[n-k] % MOD)%MOD;
	}
}