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 <cstdio>
#include <string>
#include <vector>
#include <algorithm>
#include <functional>
#include <iomanip>
#include <stdlib.h>
#include <string.h>
#include <cstring>
#include <cmath>
#include <map>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <stdio.h>
#include <tuple>
#include <bitset>
#include <cfloat>
#include <fstream>
#include <limits.h>
#include <list>
#include <math.h>
#include <random>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <numeric>
#include <utility>
#include <sstream>
#define int long long
#define rep(i, n) for (int i = 0; i < (n); i++)
#define P pair<int, int>
#define sz(x) x.size()
#define ALL(x) (x).begin(),(x).end()
#define ALLR(x) (x).rbegin(),(x).rend()
#define VE vector<int>
#define COUT(x) cout<<(x)<<endl
#define MA map<int,int>
#define SE set<int>
#define PQ priority_queue


template<class T> inline bool chmin(T& a, T b) {
	if (a > b) {
		a = b;
		return true;
	}
	return false;
}
template<class T> inline bool chmax(T& a, T b) {
	if (a < b) {
		a = b;
		return true;
	}
	return false;
}

using namespace std;
typedef long long ll;
using Graph = vector<vector<int>>;
ll gcd(ll a, ll b) {
	if (a < b)swap(a, 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;
}
bool prime(int n) {
	for (int i = 2; i <= sqrt(n); i++) {
		if (n%i == 0)return false;
	}
	return n != 1;
}


const long long MOD = 1000000007;
const long long INF = 1LL << 50;

// pair -> vector<pair<int,int>> 
//		   AandB.push_back(make_pair(a[i],b[i]))



signed main() {
	ios_base::sync_with_stdio(0);
	cin.tie(0);
	int n; cin >> n;
	VE a(n),b(n);
	int sum = 0; int sum2 = 0;
	rep(i, n) {
		cin >> a[i];
		sum += a[i];
	}
	rep(i, n) {
		cin >> b[i]; sum2 += b[i];
	}
	int tmp = 0;
	rep(i, n) {
		if (a[i] < b[i]) {
			tmp += (b[i] - a[i] + 1) / 2;
		}
		else if (a[i] > b[i]) {
			tmp += a[i] - b[i];
		}
	}
	if (tmp <= sum2 - sum)cout << "Yes" << endl;
	else cout << "No" << endl;
	return 0;
}