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;
typedef long long ll;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<ll> vl;
typedef queue<int> qi;
typedef queue<ll> ql;
typedef pair<ll,ll> pll;
typedef pair<int,ll> pil;
typedef pair<ll,int> pli;
#define pq priority_queue
#define rep(i,n) for (int i=0;i<n;i++)
#define mp make_pair
#define pb push_back
#define pf push_front
#define F first
#define S second
#define INF 200000009
#define init(x) memset(x,0,sizeof(x))
ll n,spe=0;
ll a[10005],b[10005];
int main()
{
	ios::sync_with_stdio(false);
	cin>>n;
	rep(i,n) cin>>a[i];
	rep(i,n) cin>>b[i];
	rep(i,n-1){
		if (a[i]<=b[i]) spe+=(b[i]-a[i])/2;
		else{
			if (b[i]+spe>=a[i]) spe-=a[i]-b[i];
			else{
				b[i]+=spe;
				spe=0;
				a[i+1]+=(a[i]-b[i])*2;
			}
		}
	}
	if (a[n-1]<=b[n-1]+spe) cout<<"Yes";
	else cout<<"No";
	return 0;
}