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 Rust (1.15.1) (judge verdict: Accepted)

use std::io;
use std::str::FromStr;

static MOD: u64 = 1000000007;

fn input_tuple() -> (u64, u64) {
    let mut line: String = String::new();
    io::stdin().read_line(&mut line);
    let vec: Vec<u64> = line.split_whitespace().map(|s| s.parse().unwrap()).collect();
    (vec[0], vec[1])
}

fn binary_power(x: u64, n: u64) -> u64 {
    if n == 0 {
      return 1;
    }
    if n % 2 == 0 {
      return (binary_power(x, n / 2) % MOD).pow(2) % MOD;
    } else {
      return binary_power(x, n - 1) % MOD * x % MOD;
    }
}
  
fn inverse_mod(n: u64) -> u64 {
    binary_power(n, MOD-2)
}  

fn combination(n: u64, mut m: u64) -> u64 {
    if n < m {
        return 0;
    }
    
    if m > n - m {
        m = n - m
    }

    let mut ret = 1u64;
    for i in 0..m {
        ret %= MOD;
        ret *= n - i;        
    }
    ret %= MOD;

    let mut div = 1u64;
    for i in 1..(m+1) {
        div %= MOD;
        div *= i;
    }
    div %= MOD;

    // println!("{}, {}", ret, div);
    ret * inverse_mod(div) % MOD
}

fn main() {
    let (x, y) = input_tuple();
    if x < y / 2  || y < x / 2 {
        println!("0");
        return;
    }
    if (x + y) % 3 != 0 {
        println!("0");
        return;
    }

    // println!("{}", convination((x+y)/3, (x-y).abs()/3));

    let count = (x + y) / 3;
    println!("{}", combination(count, (count + x - y) / 2));

}