#pragma gcc optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MOD = 1e9 + 7;
const int N = 1e6 + 2;

ll x, y;
ll f[N];

ll bm(ll b, ll p, ll m) {
    ll rt = 1;
    for (; p; p >>= 1, b = b * b % m)
        if (p & 1) rt = rt * b % m;
    return rt;
}

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> x >> y;
    if ((x + y) % 3) return cout << "0\n", 0;
    ll a = y - (x + y) / 3;
    ll b = x - (x + y) / 3;
    if (a < 0 || b < 0) return cout << "0\n", 0;
    f[0] = 1;
    for (ll i = 1; i <= a + b; ++i)
        f[i] = i * f[i - 1] % MOD;
    cout << f[a + b] * bm(f[a], MOD - 2, MOD) % MOD * bm(f[b], MOD - 2, MOD) % MOD << '\n';
}
### Prompt

Develop a solution in Cpp to the problem described below:

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
#pragma gcc optimize("Ofast")
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
const int MOD = 1e9 + 7;
const int N = 1e6 + 2;

ll x, y;
ll f[N];

ll bm(ll b, ll p, ll m) {
    ll rt = 1;
    for (; p; p >>= 1, b = b * b % m)
        if (p & 1) rt = rt * b % m;
    return rt;
}

int main() {
    ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    cin >> x >> y;
    if ((x + y) % 3) return cout << "0\n", 0;
    ll a = y - (x + y) / 3;
    ll b = x - (x + y) / 3;
    if (a < 0 || b < 0) return cout << "0\n", 0;
    f[0] = 1;
    for (ll i = 1; i <= a + b; ++i)
        f[i] = i * f[i - 1] % MOD;
    cout << f[a + b] * bm(f[a], MOD - 2, MOD) % MOD * bm(f[b], MOD - 2, MOD) % MOD << '\n';
}
```