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)

// ------------------------------------
// Date:2018/ 2/ 3
// Problem:AtCoder Petrozavodsk Contest b.cpp
//
// ------------------------------------

#include <iostream>
#include <string>
#include <vector>
#include <map>
#include <algorithm>
#include <random>

using namespace std;

#define FOR(i,a,b) for(int i=(int)a;i<(int)b;++i)
#define REP(i,n) FOR(i,0,n)

int main()
{
    int N,asum,bsum,n,sum;
    vector<int> A;
    vector<int> B;
    asum = bsum = sum = 0;
    cin >> N;
    REP(i,N) {
        cin >> n;
        A.push_back(n);
        asum += A[i];
    }
    REP(i,N) {
        cin >> n;
        B.push_back(n);
        bsum += B[i];
    }
    
    if (asum>bsum) {
        cout << "No" << endl;
        return 0;
    }
    
    REP(i,N) {
        n = B.at(i)-A.at(i);
        //cout << i << ":" << n << endl;
        if (n < 0) {
            sum += n;
        } else if (n >= 2){
            if (n%2==0) {
                sum += (n-2)/2;
            } else {
                sum += (n-1)/2;
            }
        }
    }
    if(sum < 0) {
        cout << "No" << endl;
    } else {
        cout << "Yes" << endl;
    }
    return 0;
}