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>
using namespace std;

int main(){
    int N;
    cin >> N;

    int a[N],b[N];
    int joken=0; //操作できる回数(Σb-Σa)
    for(int i=0;i<N;i++){
        cin >> a[i];
        joken-=a[i];
    }
    for(int i=0;i<N;i++){
        cin >> b[i];
        joken+=b[i];
    }

    int sousa=0; //必要なAの操作回数
    for(int i=0;i<N;i++){
        if(a[i]<b[i]){
            sousa+=((b[i]-a[i])/2);
            if(b[i]%2==1){sousa++;}
        }
    }

    //操作できる回数がマイナスなら絶対無理。
    //また、総操作回数が条件を上回るとA>Bとなり成立しない。
    if(joken>sousa){cout << "Yes\n";}
    else{cout << "No\n";}

    return 0;
}