#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define all(v) v.begin(), v.end()

const ll N = (ll)(2000007);
const ll MOD = (ll)(1e9 + 7);

ll f[N];
ll powerf(ll a, ll b, ll c) {
	ll res = 1;
	while(b > 0) {
		if(b&1) {
			res = (res * a);
			res = res % c;
		}
		b/=2;
		a = (a * a) % c;
	}
	return res;
}

int main() {
	
	ll ans = 0;
	f[0] = 1;
	for(int i=1; i<N; i++) {
		f[i] = f[i-1] * i;
		f[i] = f[i] % MOD;
	}

	ll x, y;
	cin >> x >> y;
	for(ll u = 0; u<=x; u++) {
		ll v = x - u;
		if(v % 2 != 0) continue;
		v/=2;
		if((2*u) + v == y) {
			ll val = (f[u] * f[v]) % MOD;
			ans = ans + (f[u+v] * powerf(val, MOD-2, MOD));
			ans = ans % MOD;
		}
	}
	cout << ans << "\n";
	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;

#define ll long long
#define all(v) v.begin(), v.end()

const ll N = (ll)(2000007);
const ll MOD = (ll)(1e9 + 7);

ll f[N];
ll powerf(ll a, ll b, ll c) {
	ll res = 1;
	while(b > 0) {
		if(b&1) {
			res = (res * a);
			res = res % c;
		}
		b/=2;
		a = (a * a) % c;
	}
	return res;
}

int main() {
	
	ll ans = 0;
	f[0] = 1;
	for(int i=1; i<N; i++) {
		f[i] = f[i-1] * i;
		f[i] = f[i] % MOD;
	}

	ll x, y;
	cin >> x >> y;
	for(ll u = 0; u<=x; u++) {
		ll v = x - u;
		if(v % 2 != 0) continue;
		v/=2;
		if((2*u) + v == y) {
			ll val = (f[u] * f[v]) % MOD;
			ans = ans + (f[u+v] * powerf(val, MOD-2, MOD));
			ans = ans % MOD;
		}
	}
	cout << ans << "\n";
	return 0;
}
```