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: Wrong Answer)

#![allow(unused_imports)]
#![allow(non_snake_case)]
use std::cmp::*;
use std::collections::*;
use std::io::Write;

#[allow(unused_macros)]
macro_rules! debug {
    ($($e:expr),*) => {
        #[cfg(debug_assertions)]
        $({
            let (e, mut err) = (stringify!($e), std::io::stderr());
            writeln!(err, "{} = {:?}", e, $e).unwrap()
        })*
    };
}

fn main() {
    let v = read_vec::<usize>();
    let (x, y) = (v[0], v[1]);
    let big_int = 1000_000_000i64 + 7;
    if (x + y) % 3 != 0 {
        println!("0");
        return;
    }

    if (max(x, y) - min(x, y)) % 2 != 0 {
        println!("0");
        return;
    }
    let diff = (max(x, y) - min(x, y)) / 2;
    if min(x, y) < diff || (min(x, y) - diff) % 3 != 0 {
        println!("0");
        return;
    }
    let same = (min(x, y) - diff) / 3;

    let fact7 = ModFactorial::new((2 * same + diff) as i64, big_int);
    let fact7 = fact7.fact;
    let inv_fact7 = (0..fact7.len())
        .map(|x| mod_inverse(fact7[x as usize], big_int))
        .collect::<Vec<_>>();
    let binom7 = ModBinominal::new((2 * same + diff) as i64, big_int, &fact7, &inv_fact7);
    println!("{}", binom7.bin[same]);
}

fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>()
        .split_whitespace()
        .map(|e| e.parse().ok().unwrap())
        .collect()
}

fn extgcd(a: i64, b: i64, x: &mut i64, y: &mut i64) -> i64 {
    let mut d = a;
    if b == 0 {
        *x = 1;
        *y = 0;
    } else {
        d = extgcd(b, a % b, y, x);
        *y -= (a / b) * (*x);
    }
    return d;
}

fn mod_inverse(a: i64, m: i64) -> i64 {
    let mut x: i64 = 0;
    let mut y: i64 = 0;
    extgcd(a, m, &mut x, &mut y);
    (m + x % m) % m
}

struct ModFactorial {
    fact: Vec<i64>,
}

// returns n! under mod m
impl ModFactorial {
    fn new(n: i64, m: i64) -> ModFactorial {
        let mut fact = vec![1; n as usize + 1];
        for i in 2..fact.len() {
            fact[i] = (i as i64 * fact[i - 1]) % m;
        }
        ModFactorial { fact: fact }
    }
}

// returns bin[i] = nCi under mod m
struct ModBinominal {
    bin: Vec<i64>,
}

impl ModBinominal {
    fn new(n: i64, m: i64, fact: &Vec<i64>, inv_fact: &Vec<i64>) -> ModBinominal {
        let mut bin = vec![1; n as usize + 1];
        if n == 0 {
            return ModBinominal { bin: bin };
        }
        for i in 1..n + 1 {
            let a2 = inv_fact[i as usize] % m;
            let a3 = inv_fact[(n - i) as usize] % m;
            bin[i as usize] = fact[n as usize] * ((a2 * a3) % m) % m;
        }
        ModBinominal { bin: bin }
    }
}