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: Accepted)

#include <iostream>
#include <string>
#include <algorithm>
#include <iomanip>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <utility>
#include <functional>
#include <deque>
#include <cctype>


using ll = long long;
typedef std::pair<int, int>p;

ll min(ll a, ll b) {
	if (a >= b)return b;
	else return a;
}
ll max(ll a, ll b) {
	if (a >= b)return a;
	else return b;
}
ll gcd(ll a, ll b) {
	if (b == 0) return a;
	return gcd(b, a%b);

}
ll lcm(ll a, ll b) {
	ll g = gcd(a, b);
	return a / g * b;
}

const ll Z = 1000000007;
const ll INF = 1 << 30;
bool flag = true;
struct edge { int to, cost; };
std::vector<edge>v[100];
bool u[100];


int main() {
	ll n;
	std::cin >> n;
	ll a[10010], b[10010], suma = 0, sumb = 0;
	for (int i = 0;i < n;i++) {
		std::cin >> a[i];
		suma += a[i];
	}
	for (int i = 0;i < n;i++) {
		std::cin >> b[i];
		sumb += b[i];
	}
	if (suma > sumb) {
		std::cout << "No" << std::endl;
		flag = false;
	}
	if (flag) {
		ll t = sumb - suma;
		for (int i =0 ;i < n;i++) {
			if (b[i] > a[i]) {
				t -= (b[i] - a[i] + 1) / 2;
			}
		}
		if (t < 0) {
			std::cout << "No" << std::endl;
		}
		else {
			std::cout << "Yes" << std::endl;
		}
	}
}