#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e6+50;
const int mod=1e9+7;

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

ll X,Y,n,m,fac[N],invfac[N];

int main(){
    fac[0]=1;
    for(int i=1;i<N;++i) fac[i]=1LL*fac[i-1]*i%mod;
    for(int i=0;i<N;++i) invfac[i]=pw(fac[i],mod-2);

    scanf("%lld%lld",&X,&Y);
    if((2*X-Y)%3!=0){
        puts("0");
        return 0;
    }
    n=(2*X-Y)/3;
    m=X-2*n;
    if(n<0 || m<0){
        puts("0");
        return 0;
    }
    ll ans=fac[n+m]*invfac[n];
    ans%=mod;
    ans*=invfac[m];
    ans%=mod;
    printf("%lld\n",ans);
    return 0;
}
### Prompt

Your task is to create 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;
typedef long long ll;
const int N=2e6+50;
const int mod=1e9+7;

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

ll X,Y,n,m,fac[N],invfac[N];

int main(){
    fac[0]=1;
    for(int i=1;i<N;++i) fac[i]=1LL*fac[i-1]*i%mod;
    for(int i=0;i<N;++i) invfac[i]=pw(fac[i],mod-2);

    scanf("%lld%lld",&X,&Y);
    if((2*X-Y)%3!=0){
        puts("0");
        return 0;
    }
    n=(2*X-Y)/3;
    m=X-2*n;
    if(n<0 || m<0){
        puts("0");
        return 0;
    }
    ll ans=fac[n+m]*invfac[n];
    ans%=mod;
    ans*=invfac[m];
    ans%=mod;
    printf("%lld\n",ans);
    return 0;
}
```