#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int P = 1e9+7;
LL f[1000001], v[1000001];
LL rp(LL now, int k)
{
    LL will = 1;
    for (; k; k >>= 1, now *= now, now %= P)
    {
        if (k & 1) will *= now, will %= P;
    }
    return will;
}
LL C(int n, int m)
{
    if(n < m) return 0;
    return f[n] * rp(f[m], P - 2) % P * rp(f[n - m], P - 2) % P;
}
void init()
{
    f[0] = 1; v[0] = 1;
    for (int i = 1; i <= 1000000; i++) //1e6以内的组合数
    {
        f[i] = f[i - 1] * i % P;
    }
}
int main(){
	int x,y;
	scanf("%d%d",&x,&y);
	init();
	if((x+y)%3){puts("0");return 0;}
    int s=(x+y)/3;
    x-=s;
    //printf("x=%d s=%d\n",x,s);
    printf("%lld\n",C(s,x));
	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 P = 1e9+7;
LL f[1000001], v[1000001];
LL rp(LL now, int k)
{
    LL will = 1;
    for (; k; k >>= 1, now *= now, now %= P)
    {
        if (k & 1) will *= now, will %= P;
    }
    return will;
}
LL C(int n, int m)
{
    if(n < m) return 0;
    return f[n] * rp(f[m], P - 2) % P * rp(f[n - m], P - 2) % P;
}
void init()
{
    f[0] = 1; v[0] = 1;
    for (int i = 1; i <= 1000000; i++) //1e6以内的组合数
    {
        f[i] = f[i - 1] * i % P;
    }
}
int main(){
	int x,y;
	scanf("%d%d",&x,&y);
	init();
	if((x+y)%3){puts("0");return 0;}
    int s=(x+y)/3;
    x-=s;
    //printf("x=%d s=%d\n",x,s);
    printf("%lld\n",C(s,x));
	return 0;
}

```