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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include<stdio.h>
long long int factorial(long long int n) {
	if (n > 0) {
		return n * factorial(n - 1) ;
	}
	else {
		return 1;
	}
}
int main(void) {
	int X, Y;
	scanf("%d %d", &X, &Y);
	if (( 2 * X -Y) % 3 != 0 || (2*Y -X) % 3!=0) {
		printf("0");
		return 0;
	}
	else {
		int a = (2*Y - X) / 3;
		int b = (2*X - Y) / 3;
		long long int n = factorial(a + b) / (factorial(a) * factorial(b));
		printf("%ld", n);
	}

	return 0;
}