#include <cstdio>
const int mo = 1e9 + 7;
using namespace std;
int x, y;
int pow(int x, int t){
	int res = 1; x %= mo;
	while (t > 0){
		if (t & 1) res = 1ll * res * x % mo;
		x = 1ll * x * x % mo;
		t >>= 1;
	}
	return res;
}
int main(){
	scanf("%d%d", &x, &y);
	if ((x + y) % 3 != 0 || (2 * y - x) % 3 != 0 || (2 * x - y) % 3 != 0 || x > 2 * y || y > 2 * x){
		puts("0");
		return 0;
	} 
	int a = (2 * y - x) / 3, b = (2 * x - y) / 3;
	int ans = 1;
	for (int i = 1; i <= (a + b); i++) ans = 1ll * ans * i % mo;
	for (int i = 1; i <= a; i++) ans = 1ll * ans * pow(i, mo - 2) % mo;
	for (int i = 1; i <= b; i++) ans = 1ll * ans * pow(i, mo - 2) % mo;
	printf("%d\n", ans);
	return 0;
}
### Prompt

Please formulate a cpp solution to the following problem:

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 <cstdio>
const int mo = 1e9 + 7;
using namespace std;
int x, y;
int pow(int x, int t){
	int res = 1; x %= mo;
	while (t > 0){
		if (t & 1) res = 1ll * res * x % mo;
		x = 1ll * x * x % mo;
		t >>= 1;
	}
	return res;
}
int main(){
	scanf("%d%d", &x, &y);
	if ((x + y) % 3 != 0 || (2 * y - x) % 3 != 0 || (2 * x - y) % 3 != 0 || x > 2 * y || y > 2 * x){
		puts("0");
		return 0;
	} 
	int a = (2 * y - x) / 3, b = (2 * x - y) / 3;
	int ans = 1;
	for (int i = 1; i <= (a + b); i++) ans = 1ll * ans * i % mo;
	for (int i = 1; i <= a; i++) ans = 1ll * ans * pow(i, mo - 2) % mo;
	for (int i = 1; i <= b; i++) ans = 1ll * ans * pow(i, mo - 2) % mo;
	printf("%d\n", ans);
	return 0;
}
```