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: Runtime Error)

import java.util.*;
public class Main {
	static int p = 1000000007;
	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		
		int x = sc.nextInt();
		int y = sc.nextInt();
		
		if((x+y)%3!=0)
			System.out.println(0);
		else{
			int m = (x+y)/3;
			int n = y-m;
			
			long[] mm = new long[m+1];
			long[] nn = new long[n+1];
			
			mm[0] = 1;
			for(int i=1;i<mm.length;i++){
				mm[i] = mm[i-1] * i % p;
			}

			nn[0] = 1;
			nn[1] = 1;
			for(int i=2;i<nn.length;i++){
				nn[i] = nn[i-1] * fp(i, p-2) % p;
			}
			
			System.out.println(mm[m] * nn[n] % p * nn[m-n] % p);
		}
	}
	static long fp(int a, int k){
		
		if(k==0)
			return 1;
		
		long num = fp(a, k/2);
		if(k % 2!=0)
			return a * num % p * num % p;
		else
			return num * num % p;
	}
}