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 <ios>
#include <iomanip>
#include <iostream>
#include <functional>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <string>
#include <sstream>
#include <complex>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <climits>
 
using namespace std;
typedef long long int ll;
typedef long double ld;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(),(x).end()
#define MOD 1000000007
#define INF (1e9)

#define PI (acos(-1))
#define print(x) cout << x << endl
 
ll gcd(ll a, ll b) { return b ? gcd(b,a%b) : a;}
ll lcm(ll a, ll b) { return a / gcd(a, b) * b; }
typedef pair <ll,ll> P;
ll nCrModp(ll n, ll r, ll p)
{
    ll C[r+1];
    memset(C, 0, sizeof(C));

    C[0] = 1;

    for (ll i = 1; i <= n; i++)
    {
        for (ll j = min(i, r); j > 0; j--)
            C[j] = (C[j] + C[j-1])%p;
    }
    return C[r];
}

int main()
{
    ll N;
    cin >> N;
    vector<ll> A(N);
    rep(i, N) cin >> A[i];
    vector<ll> B(N);
    rep(i, N) cin >> B[i];
    ll sumA = 0;
    ll sumB = 0;
    rep(i, N) {
        sumA += A[i];
        sumB += B[i];
    }
    ll cnt = 0;
    rep(i, N) {
        if (A[i] > B[i]) {
            cnt += A[i]-B[i];
        } else {
            cnt += (B[i]-A[i] + 2 - 1) / 2;
        }
    }
    if (cnt >= sumB - sumA) {
        cout << "No" << endl;
    } else {
        cout << "Yes" << endl;
    }
    return 0;
}