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 <bits/stdc++.h>
	using namespace std;
	#define IOS ios::sync_with_stdio(false), cin.tie(0)
	#define REP(i, n) for (int i = 0; i < n; i++)
	#define FOR(i, a, b) for (int i = a; i < b; i++)
	#define ALL(x) (x).begin(), (x).end()
	#define dump(x) cout << (x) << endl
	#define LMAX 9223372036854775807LL
	#define LMIN -9223372036854775807LL

	using ll = long long;
	template<typename T>
	using vec = vector<T>;
	using P = pair<ll, ll>;

	template <typename A, size_t N, typename T>
	void Fill(A (&array)[N], const T &val){
		fill((T *)array, (T *)(array + N), val);
	}
	
	const ll INF = 1e18;
	const ll mod = 1e9 + 7;

	//int dx[4] = {1,0,-1,0};
	//int dy[4] = {0,1,0,-1};
	//ll pas[51][52];
	
	signed main(){
		IOS;
		ll n;
		cin >> n;
		vec<ll> a(n),b(n);
		ll sum1,sum2;
		sum1 = sum2 = 0;
		REP(i,n){
			cin >> a[i];
			sum1 += a[i];
		}
		REP(i,n){
			cin >> b[i];
			sum2 += b[i];
		}
		ll sum = sum2 - sum1;
		ll tmp1 = 0,tmp2 = 0;
		REP(i,n){
			if(a[i] < b[i])tmp1 += ceil((b[i] - a[i]) * 0.5);
			else tmp2 += a[i] - b[i];
		}
		dump(sum >= tmp1 + tmp2 && tmp1 >= tmp2 ? "Yes" : "No");
	}