#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pll;

ll mod=1e9+7;

ll mpow(ll x, ll n) {
  ll ans=1;
  while(n>0) {
    if(n&1) {
      ans=ans*x%mod;
    }
    x=x*x%mod;
    n>>=1;
  }
  return ans;
}

vector<ll> tfact(1e7+1);
void fact(ll n) {
  for(ll i=0;i<=n;i++) {
    if(i==0) {
      tfact[i]=1;
    }
    else {
      tfact[i]=tfact[i-1]*i%mod;
    }
  }
}

ll comb(ll n, ll k) {
  return tfact.at(n)*mpow(tfact.at(k),mod-2)%mod*mpow(tfact.at(n-k),mod-2)%mod;
}

int main() {
  fact(1e7);
  ll x,y;
  cin >> x >> y;
  ll n=min(x,y)-abs(x-y);
  if(n<0||n%3) {
    cout << 0 << endl;
    return 0;
  }
  n/=3;
  cout << comb(abs(x-y)+n*2,n) << endl;
}
### Prompt

Create a solution in CPP for 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;
typedef long long ll;
typedef pair<ll,ll> pll;

ll mod=1e9+7;

ll mpow(ll x, ll n) {
  ll ans=1;
  while(n>0) {
    if(n&1) {
      ans=ans*x%mod;
    }
    x=x*x%mod;
    n>>=1;
  }
  return ans;
}

vector<ll> tfact(1e7+1);
void fact(ll n) {
  for(ll i=0;i<=n;i++) {
    if(i==0) {
      tfact[i]=1;
    }
    else {
      tfact[i]=tfact[i-1]*i%mod;
    }
  }
}

ll comb(ll n, ll k) {
  return tfact.at(n)*mpow(tfact.at(k),mod-2)%mod*mpow(tfact.at(n-k),mod-2)%mod;
}

int main() {
  fact(1e7);
  ll x,y;
  cin >> x >> y;
  ll n=min(x,y)-abs(x-y);
  if(n<0||n%3) {
    cout << 0 << endl;
    return 0;
  }
  n/=3;
  cout << comb(abs(x-y)+n*2,n) << endl;
}
```