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 <vector>
#include <set>
#include <unordered_set>
#include <map>
#include <unordered_map>
#include <algorithm>
#include <iostream>
#include <string>
#include <cstring>

using namespace std;
#define vi vector<int>
#define pii pair<int,int>
#define pb push_back
#define mp make_pair
#define all(x) x.begin(),x.end()
#define SZ(x) (int)(x.size())
#define rep(i,a,b) for(int i=a;i<b;i++)
#define per(i,a,b) for(int i=b-1;i>=a;i--)
#define inf 1000000007
#define mod 1000000007
#define ull unsigned long long
#define ll long long
template <class T,class U>inline void Max(T &a,U b){if(a<b)a=b;}
template <class T,class U>inline void Min(T &a,U b){if(a>b)a=b;}

int a[10005];
int b[10005];
int N;
int main(){
	ios_base::sync_with_stdio(false);cin.tie(NULL);
	cin >> N;
	rep(i,0,N) cin >> a[i];
	rep(i,0,N) cin >> b[i];
	ll pos = 0, neg = 0;
	rep(i,0,N) {
		int x = b[i] - a[i];
		x>0?(pos += x):(neg -= x);
	}	
	if(pos >= 2* neg) {cout << "Yes" << endl;}
	else {cout << "No" << endl;}
}