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: Runtime Error)

#include <stdio.h>
#include <math.h>

void swap(long long *xp, long long *yp) {
  long long temp = *xp; 
  *xp = *yp; 
  *yp = temp; 
} 

long long modinv(long long a, long long m) {
  long long b = m, u = 1, v = 0;
  while (b) {
    long long t = a / b;
    a -= t * b; swap(a, b);
    u -= t * v; swap(u, v);
  }
  u %= m;
  if (u < 0) u += m;
  return u;
}

int main(void) {
  int X, Y;
  scanf("%d %d", &X, &Y);
  if ((Y * 2 - X) % 3 != 0 || (X * 2 - Y) % 3 != 0) {
    printf("0\n");
    return 0;
  }

  int v = (Y * 2 - X) / 3;
  int h = (X * 2 - Y) / 3;

  long long result = 1;
  for (int i = v + 1; i <= v + h; i++) {
    result *= i;
    result %= 1000000007;
  }
  for (int i = 2; i <= h; i++) {
    result = result * modinv(i, 1000000007) % 1000000007;
  }

  printf("%lld", result);
}