// In the name of God

#include <bits/stdc++.h>

using namespace std;

const long long mod = 1e9+7;

long long fact[(int)5e7];

long long fast_pow(long long x, long long y) {
	long long ret = 1;
	for(;y;y>>=1) {
		if(y&1) ret=(ret*x)%mod;
		x=(x*x)%mod;
	}
	return ret;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	long long x, y;
	cin >> x >> y;
	if( (2*y-x)%3 || (2*x-y)%3 ) {
		cout << 0;
		exit(0);
	}
	long long a = (2*y-x)/3;
	long long b = (2*x-y)/3;
	if(a < 0 || b < 0) {
		cout << 0;
		exit(0);
	}
	fact[0] = 1;
	for(int i=1;i<=5e7;i++) fact[i]=(fact[i-1]*i)%mod;
	cout << (((fact[a+b]*fast_pow(fact[a], mod-2))%mod)*fast_pow(fact[b], mod-2))%mod;
	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
// In the name of God

#include <bits/stdc++.h>

using namespace std;

const long long mod = 1e9+7;

long long fact[(int)5e7];

long long fast_pow(long long x, long long y) {
	long long ret = 1;
	for(;y;y>>=1) {
		if(y&1) ret=(ret*x)%mod;
		x=(x*x)%mod;
	}
	return ret;
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	long long x, y;
	cin >> x >> y;
	if( (2*y-x)%3 || (2*x-y)%3 ) {
		cout << 0;
		exit(0);
	}
	long long a = (2*y-x)/3;
	long long b = (2*x-y)/3;
	if(a < 0 || b < 0) {
		cout << 0;
		exit(0);
	}
	fact[0] = 1;
	for(int i=1;i<=5e7;i++) fact[i]=(fact[i-1]*i)%mod;
	cout << (((fact[a+b]*fast_pow(fact[a], mod-2))%mod)*fast_pow(fact[b], mod-2))%mod;
	return 0;
}

```