#include<iostream>
using namespace std;
typedef long long ll;
#define MOD 1000000007

ll mul(ll n){
    ll p;
    p = 1;
    for(int i = 1;i < n+1;i++){
        p = p*i%MOD;
    }
    return p;
}

ll modinv(long long a) {
    long long b = MOD, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= MOD; 
    if (u < 0) u += MOD;
    return u;
}

int main(){
    ll x,y;
    cin >> x >> y;
    if((2*x-y)%3 == 0 && (2*y-x)%3 == 0 && 2*x-y >=0 && 2*y-x>=0){
        ll n = (2*x-y)/3;
        ll m = (2*y-x)/3;
        cout << mul(n+m)%MOD*modinv(mul(n))%MOD*modinv(mul(m))%MOD << endl;
    }else{
        cout << 0 << endl;
    }
}
### Prompt

Develop a solution in CPP 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<iostream>
using namespace std;
typedef long long ll;
#define MOD 1000000007

ll mul(ll n){
    ll p;
    p = 1;
    for(int i = 1;i < n+1;i++){
        p = p*i%MOD;
    }
    return p;
}

ll modinv(long long a) {
    long long b = MOD, u = 1, v = 0;
    while (b) {
        long long t = a / b;
        a -= t * b; swap(a, b);
        u -= t * v; swap(u, v);
    }
    u %= MOD; 
    if (u < 0) u += MOD;
    return u;
}

int main(){
    ll x,y;
    cin >> x >> y;
    if((2*x-y)%3 == 0 && (2*y-x)%3 == 0 && 2*x-y >=0 && 2*y-x>=0){
        ll n = (2*x-y)/3;
        ll m = (2*y-x)/3;
        cout << mul(n+m)%MOD*modinv(mul(n))%MOD*modinv(mul(m))%MOD << endl;
    }else{
        cout << 0 << endl;
    }
}
```