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: Accepted)

#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <string>
#include <algorithm>
#include <tuple>
#include <list>
#include <deque>
#include <stack>
#include <complex>
#include <functional>
#include <regex>
#include <map>
#include <numeric>

using namespace std;

#define REP(i,a,b) for(int i=(a); i<(b); ++(i))
#define rep(i,n) REP(i,0,n)

const int dx[] = {0, 1, 0, -1, 1, -1, 1, -1};
const int dy[] = {1, 0, -1, 0, 1, -1, -1, 1};

using ll = long long;
using ld = long double;
using ull = unsigned long long;

const int MAX = 2000000;
const int MOD = 1000000007;

ll fac[MAX], finv[MAX], inv[MAX];

void com_init() {
  fac[0] = fac[1] = 1;
  finv[0] = finv[1] = 1;
  inv[1] = 1;
  REP(i, 2, MAX) {
    fac[i] = fac[i - 1] * i % MOD;
    inv[i] = MOD - inv[MOD % i] * (MOD / i) % MOD;
    finv[i] = finv[i - 1] * inv[i] % MOD;
  }
}

ll 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(void) {
  int X, Y, S;
  cin >> X >> Y;
  S = X + Y;
  if (S % 3 != 0 || X < S / 3 || Y < S / 3) {
    cout << 0 << endl;
  } else {
    com_init();
    cout << com(S / 3, X - S / 3) << endl;
  }
  return 0;
}