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;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> iint;
typedef pair<ll,ll> llll;
#define ALL(x) (x).begin(),(x).end()
const ll zero = 0;
const ll one = 1;
const ll INF = 9223372036854775807; //10^18
const int inINF = 2147483647; //10^9
const ll MOD = 1000000007; //10^9+7
const ll MOD2 = 998244353;
void Yes() {printf("Yes\n");}
void No() {printf("No\n");}
void YES() {printf("YES\n");}
void NO() {printf("NO\n");}

int main(){
    int N;
    cin >> N;
    vector<ll> a(N), b(N);
    for (int i = 0; i < N; i++) {
        cin >> a[i];
    }
    for(int i = 0; i < N; i++) {
        cin >> b[i];
    }   
    ll ca = 0; ll cb = 0;
    ll ma = 0;
    for (int i = 0; i < N; i++) {
        ma += b[i]; ma -= a[i];
    }
    for (int i = 0; i < N; i++) {
        if(a[i] > b[i]){
            cb += a[i] - b[i];
        }
        else if(a[i] < b[i]){
            if((b[i] - a[i]) % 2 == 0){
                ca += (b[i] - a[i]) / 2;
            }
            else{
                ca += (b[i] + 1- a[i]) / 2;
                cb += 1;
            }
        }
    }
    if(ca <= ma && cb <= ma){
        Yes();
    }
    else{
        No();
    }

}