#include<bits/stdc++.h>
using namespace std;
using ll = long long;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MOD = 1e9+7, MX = 1e6+5;
int x,y,z;
ll exp(ll b, ll n){
	ll res = 1;
	for(; n; n/=2,(b*=b)%=MOD) if(n%2) (res *= b) %= MOD;
	return res;
}
ll fact[MX] = {1}, invf[MX];
void setcomb(int n){
	for(int i = 1; i <= n; i++) fact[i] = (fact[i-1] * i) % MOD;
	invf[n] =  exp(fact[n] ,MOD-2);
	for(int i = n; i > 0; i--) invf[i-1] = (invf[i] * i) % MOD;
}
inline ll comb(int a, int b){return fact[a] * invf[a-b] % MOD * invf[b] % MOD;}
int main(){
	scanf("%d%d", &x, &y);
	if((x+y)%3) return 0*puts("0");
	z = (x+y)/3;
	x -= z; y -= z;
	setcomb(z);
	printf("%lld\n", comb(z,x));
}
### 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;
using ll = long long;
#define rep(i,n) for(int i=0; i<(n); i++)
const int MOD = 1e9+7, MX = 1e6+5;
int x,y,z;
ll exp(ll b, ll n){
	ll res = 1;
	for(; n; n/=2,(b*=b)%=MOD) if(n%2) (res *= b) %= MOD;
	return res;
}
ll fact[MX] = {1}, invf[MX];
void setcomb(int n){
	for(int i = 1; i <= n; i++) fact[i] = (fact[i-1] * i) % MOD;
	invf[n] =  exp(fact[n] ,MOD-2);
	for(int i = n; i > 0; i--) invf[i-1] = (invf[i] * i) % MOD;
}
inline ll comb(int a, int b){return fact[a] * invf[a-b] % MOD * invf[b] % MOD;}
int main(){
	scanf("%d%d", &x, &y);
	if((x+y)%3) return 0*puts("0");
	z = (x+y)/3;
	x -= z; y -= z;
	setcomb(z);
	printf("%lld\n", comb(z,x));
}
```