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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <iostream>
#include <fstream>
#include <string>
#include <vector>
#include <queue>
#include <cmath>
#include <algorithm>
#include <random>
#include <chrono>
#include <iomanip>
#define lint long long int
#define FOR(x, to) for(int x=0; x<(int)(to); x++) // for簡略表記
#define DUMP(ar) for(int loop=0; loop<(int)ar.size(); loop++) cout << "[" << loop << "]:" << ar[loop] << endl // 配列一覧表示
#define DUMPL(ar) for(int loop=0; loop<(int)ar.size(); loop++) { cout << ar[loop]; if(loop<(int)ar.size()-1) cout << ' '; } cout << endl; // 配列一行一覧表示
#define COUT(x) cout << x << endl // 単要素出力
#define ALL(ar) ar.begin(), ar.end() // コンテナの全要素
#define LOOPD(i) cout << "ループ[" << i << "]回目---------------" << endl // ループ回数デバグ
using namespace std;
// cout << fixed << setprecision(4) で小数点以下4桁の桁そろえ
// cout << setw(5) << ... で表示形式を5桁に設定 5桁未満なら頭に空白がつく 5桁越なら先頭から5桁まで
// cout << setw(5) << setfill('0') << ... で空白部分の埋め文字に5を使う
// getline(cin, [変数]); で空白などを含む入力の受け取り(行ごとの受け取り)
// cin >> ws; でそこまでにストリームに残されたホワイトスペースを取り除く(=以降のgetlineの有効化)

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n;
    cin >> n;
    vector<int> a(n), b(n);
    FOR(i, n) cin >> a[i];
    FOR(i, n) cin >> b[i];

    lint a_cnt = 0, b_cnt = 0;
    for(int i=0; i<n; i++){
        lint tmp = a[i] - b[i];
        if(tmp > 0){
            b_cnt += tmp;
        }else if(tmp < 0){
            a_cnt += (-tmp) / 2;
        }
    }
    // COUT(a_cnt);
    // COUT(b_cnt);
    if(a_cnt == b_cnt || (a_cnt - b_cnt) == 1){
        COUT("Yes");
        return 0;
    }else{
        COUT("No");
        return 0;
    }
    
    return 0;
}