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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int maxx = 1e6 + 55;
const int Mod = 1e9 + 7;
LL x, y;
LL fac[maxx];

LL qwe(LL a, LL n) { //快速幂
    LL res = 1LL;
    while(n) {
        if(n & 1) res = res * a % Mod;
        a = a * a % Mod;
        n >>= 1;
    }
    return res;
}

LL C(LL n, LL m) { //组合数
	return 1LL * fac[n] * qwe(fac[n - m], Mod - 2) % Mod * qwe(fac[m], Mod - 2) % Mod;
}

int main() {
	fac[0] = 1;
    for(int i = 1; i < maxx; i++) fac[i] = 1LL * fac[i - 1] * i % Mod;
	cin >> x >> y;
	LL a = x - (x + y) / 3;
	LL b = y - (x + y) / 3;
	if((x + y) % 3) cout << 0 << endl;
	else cout << C(a + b, a) << endl;
	return 0;
}

/*
设"右1下2"进行了a次, "下1右2"转移b次
题意转化为: 有a个1与b个0进行组合, 可以制造出多少不同的串
=> ans = C(a + b, a) = C(a + b, b)
*/