#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int mod = 1000000007;

ll rui(ll a, ll b) {
  ll ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return ans;
}

ll comb(ll a, ll b) {
  ll ans = 1;
  for (ll i = a; i > a - b; i--) {
    ans = ans * i % mod;
  }
  for (ll i = 1; i <= b; i++) {
    ans = (ans * rui(i, mod - 2)) % mod;
  }
  return ans;
}

int main() {
  int X, Y;
  cin >> X >> Y;
  if ((X + Y) % 3 != 0)
    cout << 0 << endl;
  else {
    int A = (X + Y) / 3;
    X -= A;
    Y -= A;
    if (X < 0 || Y < 0)
      cout << 0 << endl;
    else {
      cout << comb(A, X) << 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;
using ll = long long;
const int mod = 1000000007;

ll rui(ll a, ll b) {
  ll ans = 1;
  while (b > 0) {
    if (b & 1) ans = ans * a % mod;
    a = a * a % mod;
    b /= 2;
  }
  return ans;
}

ll comb(ll a, ll b) {
  ll ans = 1;
  for (ll i = a; i > a - b; i--) {
    ans = ans * i % mod;
  }
  for (ll i = 1; i <= b; i++) {
    ans = (ans * rui(i, mod - 2)) % mod;
  }
  return ans;
}

int main() {
  int X, Y;
  cin >> X >> Y;
  if ((X + Y) % 3 != 0)
    cout << 0 << endl;
  else {
    int A = (X + Y) / 3;
    X -= A;
    Y -= A;
    if (X < 0 || Y < 0)
      cout << 0 << endl;
    else {
      cout << comb(A, X) << endl;
    }
  }
}

```