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 long double ld;
typedef pair<int,int> P;
typedef pair<ll, ll> Pll;
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define repn(i,n) for(int i = 0; i <= (int)(n); i++)
#define srep(i,l,n) for(int i = l; i < (int)(n); i++)
#define srepn(i,l,n) for(int i = l; i <= (int)(n); i++)
#define drep(i,n) for(int i = (int)(n-1); i >= 0; i--)
#define drepn(i,n) for(int i = (int)(l); i >= 0; i--)
#define size(s) (int)s.size()
#define debug(var)  do{std::cout << #var << " : ";view(var);}while(0)
template<typename T> void view(T e){std::cout << e << std::endl;}
template<typename T> void view(const std::vector<T>& v){for(const auto& e : v){ std::cout << e << " "; } std::cout << std::endl;}
template<typename T> void view(const std::vector<std::vector<T> >& vv){ for(const auto& v : vv){ view(v); } }
template<class T>bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T>bool chmin(T &a, const T &b) { if (b<a) { a=b; return 1; } return 0; }
void Yes(){cout << "Yes" << endl;}
void No(){cout << "No" << endl;}
void YES(){cout << "YES" << endl;}
void NO(){cout << "NO" << endl;}
void err() {cout << -1 << endl;}
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define pb push_back
#define ep emplace_back

const int MOD = 1000000007;
const int INF = 1e9;

#define PI acos(-1);
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
int ddx[8] = {1,1,1,-1,-1,-1,0,0};
int ddy[8] = {0,1,-1,0,1,-1,1,-1};


int main(){
    int n;
    cin >> n;
    vector<int> a(n), b(n);
    rep(i, n) cin >> a[i];
    rep(i, n) cin >> b[i];
    ll suma, sumb;
    suma = sumb = 0;
    ll mi = 0;
    rep(i, n) {
        suma += a[i];
        sumb += b[i];
        if(b[i] > a[i]) {
            if((b[i]&1)^(a[i]&1)) mi += (b[i] - a[i] + 1) / 2 + 1;
            else mi += (b[i] - a[i]) / 2;
        }else {
            mi += a[i] - b[i];
        }
    }
    if(sumb < suma) No();
    else if(mi <= sumb - suma) Yes();
    else No();
}