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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#define ll long long
#define MAX 5100000
const ll mod = 1000000007;
ll fac[MAX], finv[MAX], inv[MAX];

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;
    }
}
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(){
  int x, y;
  scanf("%d%d",&x,&y);
  int a = 2*x-y, b = 2*y-x;
  if(a%3 || b%3)a=-3;
  a/=3;b/=3;
  COMinit();
  printf("%lld\n", COM(a+b,a));
  return 0;
}