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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

long int calc(int x, int y)
{
	if(y>x/2) {
		y = x-y;
	}
	long int t = 1;
	for(int i=0;i<y;i++)
		t *= (x-i);
	for(int i=1;i<=y;i++)
		t /= i;
	return t % 10000000007;
}

int main(void)
{
	int x, y, xx, yy, X, Y;
	scanf("%d %d%*c", &x, &y);
	if((x+y)%3 != 0) {
		printf("0");
		return 0;
	}

	xx = (2*y-x)/3;
	yy = 2*xx;
	X = (x-xx+y-yy)/3;
	Y = (xx+yy)/3;
	printf("%d %d\n", X, Y);

	printf("%ld", calc(X+Y,Y));

	return 0;
}