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: Wrong Answer)

#![allow(non_snake_case)]
use proconio::input;
// use std::collections::HashMap;

fn main() {
    input! {
        N: usize,
        mut As: [i64; N],
        mut Bs: [i64; N],
    }

    // 前から1つずつ見ていって、
    // 一致している → OK
    // Bのほうが大きい → OK
    // Aのほうが大きい → うーん。
    // 4 1
    // 2 5

    // 1 2 3
    // 5 2 2
    // →
    // 5 2 3
    // 5 2 4
    // 後ろにどんどん押しつけていけばいいのでは。

    for i in 0..N - 1 {
        let a = As[i];
        let b = Bs[i];

        if a > b {
            // As末尾に2を足す
            let diff = a - b;
            As[N - 1] += diff * 2;
        } else if a < b {
            let diff = b - a;
            if diff == 1 {
                // 解決して次へ
                continue;
            }

            if diff % 2 == 0 {
                Bs[N - 1] += diff / 2;
                continue;
            } else {
                // 1の差にして、その後As[i]+=2, Bs[i]+=1するので
                // 結局偶数の場合と同じ
                Bs[N - 1] += diff / 2;
                continue;
            }
        }
    }

    if As[N - 1] <= Bs[N - 1] {
        println!("Yes");
    } else {
        println!("No");
    }
}