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)

#define _CRT_SECURE_NO_WARNINGS

#include<iostream>
#include<string>
#include<vector>
#include<list>
#include<stack>
#include<queue>
#include<map>
#include<tuple>
#include<algorithm>
#include<numeric>
#include<math.h>

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

using namespace std;

using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;

const int mod = 1e9 + 7;
const long double pi = 3.141592653589793238462643383279502884197;
const long double eps = 1e-7;

const int MAX = 10000;

int main() {
	int N;
	cin >> N;
	int a[MAX], b[MAX];
	int plusNum=0, minusNum=0;

	for (int i = 0; i < N; ++i) {
		cin >> a[i];
	}
	for (int i = 0; i < N; ++i) {
		cin >> b[i];
	}

	for (int i = 0; i < N; ++i) {
		if (a[i] - b[i] > 0) {
			plusNum += (a[i] - b[i]);
		}
		else if (b[i] - a[i] > 0) {
			minusNum += (b[i] - a[i])/2;
		}
	}

	if (minusNum < plusNum) cout << "No";
	else cout << "Yes";

	system("pause");
	return 0;
}