#include <bits/stdc++.h>
using namespace std;

int main(int, char**)
{
	long long X, Y;
	cin >> X >> Y;
	if ((2 * Y - X) % 3 != 0 || (2 * X - Y) % 3 != 0)
	{
		cout << 0 << endl;
		return 0;
	}
	long long alpha = (2 * Y - X) / 3, beta = (2 * X - Y) / 3;
	if (alpha < 0 || beta < 0)
	{
		cout << 0 << endl;
		return 0;
	}
	const long long M = 1000000000 + 7;
	long long n = alpha + beta;
	long long k = min(alpha, beta);
	vector<long long> fact(n + 1, 1), inv1(n + 1, 1), inv2(n + 1, 1);
	for (long long i = 2; i <= n; ++i)
	{
		fact[i] = fact[i - 1] * i % M;
		inv2[i] = M - inv2[M % i] * (M / i) % M;
		inv1[i] = inv1[i - 1] * inv2[i] % M;
	}
	cout << fact[n] * (inv1[k] * inv1[n - k] % M) % M << endl;
	return 0;
}

### Prompt

Your challenge is to write 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 <bits/stdc++.h>
using namespace std;

int main(int, char**)
{
	long long X, Y;
	cin >> X >> Y;
	if ((2 * Y - X) % 3 != 0 || (2 * X - Y) % 3 != 0)
	{
		cout << 0 << endl;
		return 0;
	}
	long long alpha = (2 * Y - X) / 3, beta = (2 * X - Y) / 3;
	if (alpha < 0 || beta < 0)
	{
		cout << 0 << endl;
		return 0;
	}
	const long long M = 1000000000 + 7;
	long long n = alpha + beta;
	long long k = min(alpha, beta);
	vector<long long> fact(n + 1, 1), inv1(n + 1, 1), inv2(n + 1, 1);
	for (long long i = 2; i <= n; ++i)
	{
		fact[i] = fact[i - 1] * i % M;
		inv2[i] = M - inv2[M % i] * (M / i) % M;
		inv1[i] = inv1[i - 1] * inv2[i] % M;
	}
	cout << fact[n] * (inv1[k] * inv1[n - k] % M) % M << endl;
	return 0;
}

```