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 "bits/stdc++.h"
using namespace std;
#ifdef _DEBUG
#include "dump.hpp"
#else
#define dump(...)
#endif

#define int long long
#define rep(i,a,b) for(int i=(a);i<(b);i++)
#define rrep(i,a,b) for(int i=(b)-1;i>=(a);i--)
#define all(c) begin(c),end(c)
const int INF = sizeof(int) == sizeof(long long) ? 0x3f3f3f3f3f3f3f3fLL : 0x3f3f3f3f;
const int MOD = (int)(1e9) + 7;
template<class T> bool chmax(T &a, const T &b) { if (a < b) { a = b; return true; } return false; }
template<class T> bool chmin(T &a, const T &b) { if (b < a) { a = b; return true; } return false; }

signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int N; cin >> N;
	vector<int> a(N); rep(i, 0, N) {
		cin >> a[i];
	}
	vector<int> b(N); rep(i, 0, N) {
		cin >> b[i];
	}
	int suma = accumulate(all(a), 0LL);
	int sumb = accumulate(all(b), 0LL);
	dump(suma, sumb);
	if (sumb < suma) {
		cout << "No" << endl;
	}
	else {
		int cnt1 = 0;
		int cnt2 = 0;
		rep(i, 0, N) {
			if (a[i] > b[i])
				cnt2 += a[i] - b[i];
			else {
				cnt1 += (b[i] - a[i]) / 2;
				if ((b[i] - a[i]) % 2) {
					cnt1++;
					cnt2++;
				}
			}
			dump(cnt1, cnt2);
		}
		dump(cnt1, cnt2);
		if (cnt1 >= cnt2) {
			cout << "Yes" << endl;
		}
		else {
			cout << "No" << endl;
		}
	}
	return 0;
}