#include <stdio.h>
long long p = 1000000007;
long long x, y, z;
long long rui(long long a,long long b) {
	if (b == 0)return 1;
	if (b == 1)return a;
	if (b == 2)return a * a % p;
	if (b % 2)return rui(rui(a, b / 2), 2) * a % p;
	else return rui(rui(a, b / 2), 2);
}
int main() {
	scanf("%lld%lld", &x, &y);
	if ((x + y) % 3) {
		printf("0\n");
		return 0;
	}
	if (x > 2 * y || y > x * 2) {
		printf("0\n");
		return 0;
	}
	z = (x + y) / 3;
	x -= z;
	y -= z;
	z = 1;
	for (long long i = 1; i <= x + y; i++) {
		z *= i; z %= p;
	}
	for (long long i = 1; i <= x; i++) {
		z *= rui(i, p - 2);
		z %= p;
	}
	for (long long i = 1; i <= y; i++) {
		z *= rui(i, p - 2);
		z %= p;
	}
	printf("%lld\n", z);
}

### Prompt

Develop a solution in CPP to the problem described below:

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.

Examples

Input

3 3


Output

2


Input

2 2


Output

0


Input

999999 999999


Output

151840682

### Response

```cpp
#include <stdio.h>
long long p = 1000000007;
long long x, y, z;
long long rui(long long a,long long b) {
	if (b == 0)return 1;
	if (b == 1)return a;
	if (b == 2)return a * a % p;
	if (b % 2)return rui(rui(a, b / 2), 2) * a % p;
	else return rui(rui(a, b / 2), 2);
}
int main() {
	scanf("%lld%lld", &x, &y);
	if ((x + y) % 3) {
		printf("0\n");
		return 0;
	}
	if (x > 2 * y || y > x * 2) {
		printf("0\n");
		return 0;
	}
	z = (x + y) / 3;
	x -= z;
	y -= z;
	z = 1;
	for (long long i = 1; i <= x + y; i++) {
		z *= i; z %= p;
	}
	for (long long i = 1; i <= x; i++) {
		z *= rui(i, p - 2);
		z %= p;
	}
	for (long long i = 1; i <= y; i++) {
		z *= rui(i, p - 2);
		z %= p;
	}
	printf("%lld\n", z);
}

```