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: Wrong Answer)

#include <bits/stdc++.h>
#define int long long
using namespace std;


/* エイリアス */
#define ll long long
#define ld long double
#define vi vector<int>
#define vll vector<ll>
#define pii pair<int,int>
#define pll pair<ll,ll>
#define vpi vector<pii>
#define vpll vector<pll>
#define endl '\n'

#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define em emplace
#define mp make_pair


/* 関数マクロ */
#define rep(i, n) for (ll i = 0; i < n; ++i)
#define rep2(i, n, m) for (ll i = n; i <= m; ++i)
#define rep3(i, n, m) for (ll i = n; i >= m; --i)
#define all(v) v.begin(), v.end()
#define si(v) int(v.size())
#define UNIQUE(v) sort(all(v)), v.erase(unique(all(v)),v.end())


/* 定数 */
const ll mod = 1e9 + 7;
const ll infll = (1LL << 62) - 1;
const ll inf = (1LL << 30) - 1;


/* その他 */
template<class S, class T> inline bool chmax(S &a, T b) { if (a < b) { a = b; return 1; } return 0; }
template<class S, class T> inline bool chmin(S &a, T b) { if (a > b) { a = b; return 1; } return 0; }
template<class T> using pq = priority_queue<T>;
template<class T> using pqg = priority_queue<T, vector<T>, greater<T>>;


signed main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cout << fixed << setprecision(15);

  int n; cin >> n;
  vi a(n), b(n);
  int plus = 0, minus = 0, eqq = 0;

  rep(i, n) cin >> a[i];
  rep(i, n) cin >> b[i];

  rep(i, n) {
    int d = b[i] - a[i];

    if (d < 0) minus += abs(d);
    else if (d > 0) {
      if (d % 2) ++minus;
      plus += (d + 1) / 2;
    } else ++eqq;
  }

  if (minus > plus) {
    cout << "No" << endl;
    return 0;
  }

  if (minus && plus - eqq <= minus) {
    cout << "Yes" << endl;
    return 0;
  }

  if (plus != minus) cout << "No" << endl;
  else cout << "Yes" << endl;
}