#include <bits/stdc++.h>
using namespace std;
int64_t MOD = 1000000007;

int64_t modinv(int64_t a, int64_t m)
{
  int64_t b = m, u = 1, v = 0;
  while (b)
  {
    int64_t t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0)
    u += m;
  return u;
}

int64_t nCk(int64_t n, int64_t k)
{
  int64_t res = 1;
  for (int64_t i = 1; i <= k; i++)
  {
    res = ((res * (n - i + 1)) % MOD * modinv(i, MOD)) % MOD;
  }
  return res;
}

int main()
{
  int64_t X, Y, ans;
  cin >> X >> Y;
  if (Y > 2 * X || 2 * Y < X || (X + Y) % 3 != 0)
  {
    cout << 0 << endl;
    return 0;
  }
  ans = nCk((X + Y) / 3, X - (X + Y) / 3);
  cout << ans % MOD << endl;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int64_t MOD = 1000000007;

int64_t modinv(int64_t a, int64_t m)
{
  int64_t b = m, u = 1, v = 0;
  while (b)
  {
    int64_t t = a / b;
    a -= t * b;
    swap(a, b);
    u -= t * v;
    swap(u, v);
  }
  u %= m;
  if (u < 0)
    u += m;
  return u;
}

int64_t nCk(int64_t n, int64_t k)
{
  int64_t res = 1;
  for (int64_t i = 1; i <= k; i++)
  {
    res = ((res * (n - i + 1)) % MOD * modinv(i, MOD)) % MOD;
  }
  return res;
}

int main()
{
  int64_t X, Y, ans;
  cin >> X >> Y;
  if (Y > 2 * X || 2 * Y < X || (X + Y) % 3 != 0)
  {
    cout << 0 << endl;
    return 0;
  }
  ans = nCk((X + Y) / 3, X - (X + Y) / 3);
  cout << ans % MOD << endl;
}

```