Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

3 3

Sample Output 1

2

There are two ways: (0,0) \to (1,2) \to (3,3) and (0,0) \to (2,1) \to (3,3).

Sample Input 2

2 2

Sample Output 2

0

The knight cannot reach (2,2).

Sample Input 3

999999 999999

Sample Output 3

151840682

Print the number of ways modulo 10^9 + 7.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

// copied by Nurstan Duisengaliev
// skatal
#include <bits/stdc++.h>
 
#define ll long long
#define all(x) x.begin(), x.end()
#define in insert
#define mp make_pair
#define F first
#define S second
#define ppf pop_front
#define pb push_back
#define ppb pop_back
#define pf push_front
#define pii pair <int, int>
#define boost() ios_base::sync_with_stdio(0), cin.tie(0)
#define sz(x) x.size()
#define int ll 
using namespace std;
 
const int N = (int)2e6 + 123;
const int mod = (int)1e9 + 7;
const ll INF = (ll)1e18 + 7;
ll fac[N];
ll binpow (int x, int y) {
	if (y == 0) return 1;
	if (y % 2 == 0) {
		ll o = binpow (x, y / 2);
		return (o * o) % mod;
	}	
	else {
		return ((binpow (x, y - 1) * x) % mod);
	}
}
ll cnk (int n, int k) {
	ll x = fac[n];
	ll y = (fac[k] * fac[n - k]) % mod;
	return (x * binpow (y, mod - 2)) % mod;	
}
void solve () {
	int n, m;
	cin >> n >> m;
	fac[0] = 1;
	for (int i = 1; i <= (int)1e6; i ++) {
		fac[i] = fac[i - 1] * i;
		fac[i] %= mod;
	} 
	ll ans = 0;
	for (int i = 1; i <= n; i ++) {
		int x = i, y = 2 * i;
		if (x <= n && y <= m) {
			int b = m - y;
			if (2 * b + x == n) {
				ans += cnk (i + b, b);    
				ans %= mod;
			}
		}			
	}
	cout << ans;
}
 
main () {
//	freopen (".in", "r", stdin);
//	freopen (".out", "w", stdout);
	boost ();
	int TT = 1;
//	cin >> TT;
	while (TT --) {
		solve ();
	}
	return 0;	
}