#include <bits/stdc++.h>
#define ll long long
#define ull unsigned long long
#define fi first
#define se second
using namespace std;
typedef pair<int,ll> P;
const int MOD=1e9+7;
int X,Y;
int A(int x)
{
    int re=1;
    for(int i=1;i<=x;i++)
        re=(ll)re*i%MOD;
    return re;
}
int ksm(int x,int y)
{
    int re=1;
    while(y)
    {
        if(y&1) re=(ll)re*x%MOD;
        x=(ll)x*x%MOD;
        y>>=1;
    }
    return re;
}
int inv(int x)
{
    return ksm(x,MOD-2);
}
int main()
{
    cin>>X>>Y;
    if((2*X-Y)%3!=0)
        return printf("0\n")*0;
    int n=(2*X-Y)/3,m=X-2*n;
    if(n<0 || m<0)
        return printf("0\n")*0;
    printf("%lld\n",(ll)A(m+n)*inv(A(m))%MOD*inv(A(n))%MOD);
    return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
#define ll long long
#define ull unsigned long long
#define fi first
#define se second
using namespace std;
typedef pair<int,ll> P;
const int MOD=1e9+7;
int X,Y;
int A(int x)
{
    int re=1;
    for(int i=1;i<=x;i++)
        re=(ll)re*i%MOD;
    return re;
}
int ksm(int x,int y)
{
    int re=1;
    while(y)
    {
        if(y&1) re=(ll)re*x%MOD;
        x=(ll)x*x%MOD;
        y>>=1;
    }
    return re;
}
int inv(int x)
{
    return ksm(x,MOD-2);
}
int main()
{
    cin>>X>>Y;
    if((2*X-Y)%3!=0)
        return printf("0\n")*0;
    int n=(2*X-Y)/3,m=X-2*n;
    if(n<0 || m<0)
        return printf("0\n")*0;
    printf("%lld\n",(ll)A(m+n)*inv(A(m))%MOD*inv(A(n))%MOD);
    return 0;
}

```