#include <bits/stdc++.h>

using namespace std;
typedef long long int lli;

const int MOD = 1000000007;

int X, Y;

lli inv(lli k)
{
	if( k == 1 ) return 1;

	lli val = MOD - MOD/k;
	val *= inv( MOD%k );

	return val%MOD;
}

int main()
{
	scanf("%d %d",&X,&Y);

	int A, B;

	A = (2*Y - X)/3;
	B = (2*X - Y)/3;

	if( (2*Y - X)%3 != 0 || (2*X - Y)%3 != 0 || A < 0 || B < 0 )
	{
		printf("0\n");
		return 0;
	}

	lli ans = 1;

	for(int i = 1 ; i <= A + B ; i++)
	{
		ans *= i;
		ans %= MOD;
	}

	for(int i = 0 ; i < 2 ; i++)
	{
		lli fat = 1;

		for(int j = 1 ; j <= A ; j++)
		{
			fat *= j;
			fat %= MOD;
		}

		ans *= inv( fat );
		ans %= MOD;
		
		swap( A , B );
	}

	printf("%lld\n",ans);
}
### Prompt

Create a solution in Cpp for 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 <bits/stdc++.h>

using namespace std;
typedef long long int lli;

const int MOD = 1000000007;

int X, Y;

lli inv(lli k)
{
	if( k == 1 ) return 1;

	lli val = MOD - MOD/k;
	val *= inv( MOD%k );

	return val%MOD;
}

int main()
{
	scanf("%d %d",&X,&Y);

	int A, B;

	A = (2*Y - X)/3;
	B = (2*X - Y)/3;

	if( (2*Y - X)%3 != 0 || (2*X - Y)%3 != 0 || A < 0 || B < 0 )
	{
		printf("0\n");
		return 0;
	}

	lli ans = 1;

	for(int i = 1 ; i <= A + B ; i++)
	{
		ans *= i;
		ans %= MOD;
	}

	for(int i = 0 ; i < 2 ; i++)
	{
		lli fat = 1;

		for(int j = 1 ; j <= A ; j++)
		{
			fat *= j;
			fat %= MOD;
		}

		ans *= inv( fat );
		ans %= MOD;
		
		swap( A , B );
	}

	printf("%lld\n",ans);
}
```