#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const ll mod = 1e9 + 7;
ll P(ll x, ll y)
{
	if (y % 2)
		return x * P(x, y - 1) % mod;
	if (y)
		return P(x * x % mod, y / 2);
	return 1;
}
ll M(ll x, ll y)
{
	return x * y % mod;
}
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	ll x, y;
	cin >> x >> y;
	if ((x + y) % 3)
	{
		cout << 0;
		return 0;
	}
	int z = (x + y) / 3;
    if (x < z || y < z || x > z * 2 || y > z * 2)
	{
		cout << 0;
		return 0;
	}
	int a = x - z;
	vector <int> c(a), d(a);
	iota(c.begin(), c.end(), z - a + 1);
	iota(d.begin(), d.end(), 1);
	cout << accumulate(c.begin(), c.end(), 1, M) * P(accumulate(d.begin(), d.end(), 1, M), mod - 2) % mod;
}
### 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;
using ll = long long;
const ll mod = 1e9 + 7;
ll P(ll x, ll y)
{
	if (y % 2)
		return x * P(x, y - 1) % mod;
	if (y)
		return P(x * x % mod, y / 2);
	return 1;
}
ll M(ll x, ll y)
{
	return x * y % mod;
}
int main()
{
	ios::sync_with_stdio(0);
	cin.tie(0);
	cout.tie(0);
	ll x, y;
	cin >> x >> y;
	if ((x + y) % 3)
	{
		cout << 0;
		return 0;
	}
	int z = (x + y) / 3;
    if (x < z || y < z || x > z * 2 || y > z * 2)
	{
		cout << 0;
		return 0;
	}
	int a = x - z;
	vector <int> c(a), d(a);
	iota(c.begin(), c.end(), z - a + 1);
	iota(d.begin(), d.end(), 1);
	cout << accumulate(c.begin(), c.end(), 1, M) * P(accumulate(d.begin(), d.end(), 1, M), mod - 2) % mod;
}
```