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: Accepted)

#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define repr(i, a, b) for(int i = a; i < b; i++)
#define  all(x) (x).begin(),(x).end()     // 昇順ソート
#define  rall(v) (v).rbegin(), (v).rend() // 降順ソート
#define  FastIO ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0)
typedef long long ll;
using P = pair<int,int>;
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 (a>b) { a=b; return true; } return false; }

int main(){
    int n;
    cin >> n;
    vector<ll> a(n), b(n);
    rep(i,n) cin >> a[i];
    rep(i,n) cin >> b[i];
    ll sumA = 0, sumB = 0, bCnt = 0, aCnt = 0;
    rep(i,n){
        sumA += a[i];
        sumB += b[i];
        if (a[i] > b[i]) bCnt += a[i] - b[i];
        else if (a[i] < b[i]) aCnt += ceil(((double)b[i]- a[i]) / 2);
    } 
    if (sumA > sumB){
        cout << "No" << endl;
    }
    else if (sumB - sumA >= bCnt && sumB - sumA >= aCnt){
        cout << "Yes" << endl;
    }else{
        cout << "No" << endl;
    }
    return 0;
}