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

const long mod = 1000000007;
int x, y;

long f(long n) {
  long ret = 1;
  for (long i = 1; i <= n; ++i) {
    ret = ret * i % mod;
  }
  return ret;
}

long pow(long b, long e) {
  if (e == 0) return 1;
  if (e % 2 == 1) return pow(b, e - 1) * b % mod;
  long t = pow(b, e / 2);
  return t * t % mod;
}

long inv(long n) {
  return pow(n, mod - 2);
}

long sub(long a, long b) {
  return f(a + b) * inv(f(a) * f(b) % mod) % mod;
}

int main() {
  cin >> x >> y;
  int i = 2 * x - y;
  int j = 2 * y - x;
  if (i < 0 || j < 0 || i % 3 != 0 || j % 3 != 0) {
    cout << 0 << endl;
  } else {
    cout << sub(i / 3, j / 3) << 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;

const long mod = 1000000007;
int x, y;

long f(long n) {
  long ret = 1;
  for (long i = 1; i <= n; ++i) {
    ret = ret * i % mod;
  }
  return ret;
}

long pow(long b, long e) {
  if (e == 0) return 1;
  if (e % 2 == 1) return pow(b, e - 1) * b % mod;
  long t = pow(b, e / 2);
  return t * t % mod;
}

long inv(long n) {
  return pow(n, mod - 2);
}

long sub(long a, long b) {
  return f(a + b) * inv(f(a) * f(b) % mod) % mod;
}

int main() {
  cin >> x >> y;
  int i = 2 * x - y;
  int j = 2 * y - x;
  if (i < 0 || j < 0 || i % 3 != 0 || j % 3 != 0) {
    cout << 0 << endl;
  } else {
    cout << sub(i / 3, j / 3) << endl;
  }
}
```