Problem:
Score : 300 points

Problem Statement

You are given two integer sequences of length N: a_1,a_2,..,a_N and b_1,b_2,..,b_N.
Determine if we can repeat the following operation zero or more times so that the sequences a and b become equal.

Operation: Choose two integers i and j (possibly the same) between 1 and N (inclusive), then perform the following two actions simultaneously:

Add 2 to a_i.

Add 1 to b_j.

Constraints

1 ≤ N ≤ 10 000

0 ≤ a_i,b_i ≤ 10^9 (1 ≤ i ≤ N)

All input values are integers.

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 .. a_N
b_1 b_2 .. b_N

Output

If we can repeat the operation zero or more times so that the sequences a and b become equal, print Yes; otherwise, print No.

Sample Input 1

3
1 2 3
5 2 2

Sample Output 1

Yes

For example, we can perform three operations as follows to do our job:

First operation: i=1 and j=2. Now we have a = \{3,2,3\}, b = \{5,3,2\}.

Second operation: i=1 and j=2. Now we have a = \{5,2,3\}, b = \{5,4,2\}.

Third operation: i=2 and j=3. Now we have a = \{5,4,3\}, b = \{5,4,3\}.

Sample Input 2

5
3 1 4 1 5
2 7 1 8 2

Sample Output 2

No

Sample Input 3

5
2 7 1 8 2
3 1 4 1 5

Sample Output 3

No

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

use std::str::FromStr;
use std::fmt::Debug;
use std::cmp;
#[allow(dead_code)]
fn read_as_vec<T>() -> Vec<T>
where T: FromStr, <T as FromStr>::Err : Debug{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().split_whitespace().map(|c| T::from_str(c).unwrap()).collect()
}

#[allow(dead_code)]
fn read_as_string() -> String{
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).unwrap();
    s.trim().to_string()
}

fn main(){
    let n: usize = read_as_vec()[0];
    let mut a = read_as_vec::<i64>();
    let mut b = read_as_vec::<i64>();

    if a.iter().sum::<i64>() > b.iter().sum::<i64>(){
        println!("No");
        return;
    }

    let mut ops = b.iter().sum::<i64>() - a.iter().sum::<i64>();
    let (mut aops, mut bops) = (0, 0);
    for i in 0..n{
        if a[i] < b[i]{
            let c = (b[i]-a[i]+1)/2;
            aops += c;
            if (b[i]-a[i])%2 == 0{
                a[i] = b[i];
            }else{
                b[i] += 1;
                bops += 2;

                a[i] = b[i];
            }
        }else if b[i] < a[i]{
            bops += a[i] - b[i];
            b[i] = a[i];
        }
    }

    if aops <= ops && bops <= ops && (ops-aops)*2 == (ops-bops){
        println!("Yes");
    }else{
        println!("No");
    }
}