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++ (GCC 9.2.1) (judge verdict: Wrong Answer)

#include<iostream>
#include<stdio.h>
#include<vector>
#include<algorithm>
#include<set>
#include<string>
#include<map>
#include<string.h>
#include<complex>
#include<math.h>
#include<queue>
#include <functional>
#include<time.h>
#include <stack>
#include<iomanip>
using namespace std;
#define rep(i,a,n) for(int i=(a);i<(n);i++)
#define ll long long
#define llint long long int
#define reverse(v)  reverse(v.begin(), v.end());
#define Yes(ans) if(ans)cout<<"Yes"<<endl; else cout<<"No"<<endl;
#define YES(ans) if(ans)cout<<"YES"<<endl; else cout<<"NO"<<endl;
#define hei(a) vector<a>
#define whei(a) vector<vector<a>>
#define UF  UnionFind 
#define Pint pair<int,int>
#define Pll pair<llint,llint>
#define keta(a)  fixed << setprecision(a)
constexpr auto mod = 1000000007;
//辞書順はnext_permutation( begin( v ), end( v ) );やで！
const llint INF = 100000000;
int main() {
	int n;
	cin >> n;
	hei(int) a(n), b(n);
	rep(i, 0, n)cin >> a[i];
	rep(i, 0, n)cin >> b[i];
	int m = 0, p = 0;
	rep(i, 0, n) {
		if (b[i] < a[i])m += a[i] - b[i];
		else p += b[i] - a[i];
	}
	Yes(p>=(2*m))
	return 0;
}