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;
using vi = vector<int>;
using vb = vector<bool>;
using vl = vector<long>;
using vs = vector<string>;
using vvi = vector<vector<int>>;
using vvb = vector<vector<bool>>;
using vvc = vector<vector<char>>;
using vvl = vector<vector<long>>;
using pii = pair<int, int>;
using pil = pair<int, long>;
using pll = pair<long, long>;
using vc = vector<char>;
#define fix20 cout << fixed << setprecision(20)
#define YES cout << "Yes" << endl
#define NO cout << "No" << endl
#define rep(i,n) for(int i=0; i<(int)(n);i++)
#define REP(i,s,t) for(int i=s; i<t; i++)
#define RNG(i,s,t,u) for(int i=s; i<t; i+=u)
#define MOD 1000000007
#define all(vec) vec.begin(), vec.end()

const int MAX = 510000;
long long fac[MAX], finv[MAX], inv[MAX];
// テーブルを作る前処理
void COMinit() {
    fac[0] = fac[1] = 1;
    finv[0] = finv[1] = 1;
    inv[1] = 1;
    for (int i = 2; i < MAX; i++){
        fac[i] = fac[i - 1] * i % MOD;
        inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
        finv[i] = finv[i - 1] * inv[i] % MOD;
    }
}
// 二項係数計算
long long COM(int n, int k){
    if (n < k) return 0;
    if (n < 0 || k < 0) return 0;
    return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}

int main(){
  long x,y;
  cin >> x >> y;
  COMinit();
  if((x + y) % 3 != 0) cout << 0 << endl;
  else{
    long m = (x + y)/3;
    long a = x - m;
    long b = y - m;
    cout << COM(a+b,a) << endl;
  }
}