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: Runtime Error)

#include<stdio.h>
long long int k[5000006];
long long int modpow(long long int a, long long int n, long long int p)
{
	if (n < 0)
		return 0;
	long long int res = 1;
	while (n > 0)
	{
		if (n % 2 > 0)
			res = res * a % p;
		n /= 2;
		a = a * a % p;
	}
	return res;
}
int main()
{
	int x, y;
	scanf("%d %d", &x, &y);
	long long int p = 1000000007;
	int a, b;
	if ((2 * y - x) % 3 > 0)
	{
		printf("0\n");
		return 0;
	}
	a = (2 * y - x) / 3;
	if ((2 * x - y) % 3 > 0)
	{
		printf("0\n");
		return 0;
	}
	b = (2 * x - y) / 3;
	long long int i;
	k[0] = 1;
	for (i = 1; i < 5000006; i++)
		k[i] = k[i - 1] * i % p;
	long long int ans = k[a + b];
	ans = ans * modpow(k[a], p - 2, p) % p;
	ans = ans * modpow(k[b], p - 2, p) % p;
	printf("%lld\n", ans);
	return 0;
}