#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
ll mod = 1e9+7;
ll binaryExp(ll a, ll b)
{
	ll ans = 1, x = a;
	while(b>0)
	{
		if(b&1)
			ans = (ans*x)%mod;
		x = (x*x)%mod;
		b>>=1;
	}
	return ans;
}
ll invfac(ll n)
{
	return binaryExp(n,mod-2);
}
ll nCr(ll n, ll r)
{
	vector<ll> fact(n+1,1);
	for(ll i = 2; i<=n; i++)
		fact[i] = (i*fact[i-1])%mod;
	return (((fact[n]*invfac(fact[n-r]))%mod)*invfac(fact[r]))%mod;
}

int main()
{
	ll x,y; cin>>x>>y;
	if((x+y)%3!=0)
		cout<<0<<endl;
	else
	{
		
		ll k = (x+y)/3;
		if(k>y||y>2*k)
		{
			cout<<0<<endl;
			return 0;
		}
		ll m = y - k;
		ll n = 2*k - y;
		cout<<nCr(m+n,min(m,n))<<endl;
	}
	return 0;
}

### Prompt

Your challenge is to write 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 int ll;
ll mod = 1e9+7;
ll binaryExp(ll a, ll b)
{
	ll ans = 1, x = a;
	while(b>0)
	{
		if(b&1)
			ans = (ans*x)%mod;
		x = (x*x)%mod;
		b>>=1;
	}
	return ans;
}
ll invfac(ll n)
{
	return binaryExp(n,mod-2);
}
ll nCr(ll n, ll r)
{
	vector<ll> fact(n+1,1);
	for(ll i = 2; i<=n; i++)
		fact[i] = (i*fact[i-1])%mod;
	return (((fact[n]*invfac(fact[n-r]))%mod)*invfac(fact[r]))%mod;
}

int main()
{
	ll x,y; cin>>x>>y;
	if((x+y)%3!=0)
		cout<<0<<endl;
	else
	{
		
		ll k = (x+y)/3;
		if(k>y||y>2*k)
		{
			cout<<0<<endl;
			return 0;
		}
		ll m = y - k;
		ll n = 2*k - y;
		cout<<nCr(m+n,min(m,n))<<endl;
	}
	return 0;
}

```