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 (1.42.0) (judge verdict: Runtime Error)

#![allow(unused_imports)]
use std::cmp::*;
use std::collections::*;
use itertools::Itertools;
use num::clamp;
use proconio::{input, marker::*, fastout};
use superslice::*;

fn main() {
    input! {
        n: i64
    }

    println!("0");
    input! {
        mut s0: String
    }
    if &s0 == "Vacant" {
        return;
    }
    
    println!("{}", n-1);
    input! {
        mut sn: String
    }
    if &sn == "Vacant" {
        return;
    }
    

    let mut prev = 0;
    let mut head = 1;
    let mut tail = n-1;
    let mut prev_s = s0;
    for _ in 1..20 {
        let now = (head + tail) / 2;
        println!("{}", now);
        input! {
            s: String
        }
        if &s == "Vacant" {
            return;
        }

        if ((now - prev) %2 == 0) == (s == prev_s) {
            head = now + 1;
        } else {
            tail = now - 1;
        }
        prev = now;
        prev_s = s;
    }
}