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>
using ll = long long;
#ifdef LOCAL
#define debug(...) {__VA_ARGS__}
#else
#define debug(...) 
#endif
using namespace std;
//}}}

constexpr int nax = 1e4;

int n, tab[nax];

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

    cin >> n;

    for (int i = 0; i < n; ++i)
        cin >> tab[i];
    for (int i = 0; i < n; ++i)
    {
        int v; cin >> v;
        tab[i] -= v;
    }
    ll sum = 0;
    for (int i = 0; i < n; ++i)
        sum += tab[i];

    ll op1 = -sum;
    ll op2 = -sum;

    debug(cerr << sum << '\n';);

    for (int i = 0; i < n; ++i)
    {
        if (tab[i] < 0)
        {
            op1 -= abs((tab[i] - 1) / 2);
            op2 -= (tab[i] & 1);
        }
        if (tab[i] > 0)
            op2 -= tab[i];
        debug(cerr << op1 << ' ' << op2 << '\n';);
    }
    if (op1 < 0 or op2 < 0 or op1 * 2 != op2)
        cout << "No" << '\n';
    else
        cout << "Yes" << '\n';
}