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<algorithm>
#include<vector>
#include<string>
#define LL long long
int N;
std::vector<LL> a;
std::vector<LL> b;
int main() {
	std::cin >> N;
	for (int i = 0; i < N; i++) {
		LL tmp;
		std::cin >> tmp;
		a.push_back(tmp);
	}
	for (int i = 0; i < N; i++) {
		LL tmp;
		std::cin >> tmp;
		b.push_back(tmp);
	}
	LL adda = 0;
	LL addb = 0;
	for (int i = 0; i < N - 1; i++) {
		if (a[i] - b[i] < 0) {
			int ne = (b[i] - a[i] + 1) / 2 * 2;
			if (adda >= ne) {
				adda -= ne;
				ne = 0;
			}
			else{
				ne -= adda;
				adda = 0;
			}
			addb += ne / 2;
			if ((b[i] - a[i]) % 2 != 0)addb--;
		}
		if (b[i] - a[i] < 0) {
			int  ne = a[i] - b[i];
			if (addb >= ne) {
				addb -= ne;
				ne = 0;
			}
			else {
				ne -= addb;
				addb = 0;
			}
			adda += ne * 2;
		}
	}
	if (a[N - 1] + adda <= b[N - 1] + addb)std::cout << "Yes" << std::endl;
	else std::cout << "No" << std::endl;
	return 0;
}