#include <bits/stdc++.h>
using namespace std;
#define rep(i,a,b) for(int i=int(a);i<int(b);i++)
typedef long long ll;

const int MAX = 1000010;
const ll MOD = 1000000007;

ll fac[MAX], finv[MAX], inv[MAX];

void init() {
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[1]=1;
    for(int i=2;i<MAX;i++){
        fac[i]=fac[i-1]*i%MOD;
        inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
        finv[i]=finv[i-1]*inv[i]%MOD;
    }
}

int main(){
    ll X,Y;cin>>X>>Y;
    if(X>Y)swap(X,Y);
    ll nn=Y-X,sq;
    if((X-nn)%3!=0||(X-nn)<0){
        cout<<0<<endl;
        return 0;
    }
    sq=(X-nn)/3;
    init();
    ll ans=(((fac[2*sq+nn]*finv[sq])%MOD)*finv[sq+nn])%MOD;

    cout<<ans<<endl;
}






### Prompt

Generate 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;
#define rep(i,a,b) for(int i=int(a);i<int(b);i++)
typedef long long ll;

const int MAX = 1000010;
const ll MOD = 1000000007;

ll fac[MAX], finv[MAX], inv[MAX];

void init() {
    fac[0]=fac[1]=1;
    finv[0]=finv[1]=1;
    inv[1]=1;
    for(int i=2;i<MAX;i++){
        fac[i]=fac[i-1]*i%MOD;
        inv[i]=MOD-inv[MOD%i]*(MOD/i)%MOD;
        finv[i]=finv[i-1]*inv[i]%MOD;
    }
}

int main(){
    ll X,Y;cin>>X>>Y;
    if(X>Y)swap(X,Y);
    ll nn=Y-X,sq;
    if((X-nn)%3!=0||(X-nn)<0){
        cout<<0<<endl;
        return 0;
    }
    sq=(X-nn)/3;
    init();
    ll ans=(((fac[2*sq+nn]*finv[sq])%MOD)*finv[sq+nn])%MOD;

    cout<<ans<<endl;
}






```