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: Accepted)

#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; i < n; ++i)
#define REPR(i, n) for (int i = n - 1; i >= 0; --i)
#define FOR(i, a, b) for (int i = a; i < b; ++i)
#define FORR(i, a, b) for (int i = b - 1; i >= a; --i)
#define SORT(v) sort(v.begin(), v.end())
#define SORTR(v) sort(v.rbegin(), v.rend())
#define REV(v) reverse(v.begin(), v.end())
#define ITER(itr, v) for (auto itr = v.begin(); itr != v.end(); ++itr)
#define LB(v, x) (lower_bound(v.begin(), v.end(), x) - v.begin())
#define UB(v, x) (upper_bound(v.begin(), v.end(), x) - v.begin())
#define SZ(v) (int)v.size()
using namespace std;
using ll = long long;
using P = pair<int, int>;

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);

    int N;
    cin >> N;
    vector<ll> a(N), b(N);
    ll sum_a = 0, sum_b = 0;
    REP(i, N) {
        cin >> a[i];
        sum_a += a[i];
    }
    REP(i, N) {
        cin >> b[i];
        sum_b += b[i];
    }
    ll cnt = 0;
    REP(i, N) {
        if (a[i] < b[i]) { cnt += (b[i] - a[i] + 1) / 2; }
    }
    cout << (cnt <= sum_b - sum_a ? "Yes" : "No") << endl;

    return 0;
}