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)

// ------------------------------------
// Date:2018/ 3/15
// Problem:APC 001 TwoArrays re_b.cpp
//
// ------------------------------------

#include <bits/stdc++.h>

using namespace std;

#define EACH(i,a) for (auto&& i : a)
#define FOR(i,a,b) for(int i=(int)a;i<(int)b;++i)
#define RFOR(i,a,b) for(int i=(int)b-1;i>=(int)a;--i)
#define REP(i,n) FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define ALL(a) (a).begin(),(a).end()
#define debug(x) cerr << #x << ":" << x << endl

typedef long long ll;

static const int MOD = 1000000007;

int main()
{
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  int N;
  cin >> N;
  vector< int > a(N);
  vector< int > b(N);
  REP(i, N) {
    cin >> a[i];
  }
  REP(i, N) {
    cin >> b[i];
  }
  
  ll cost = 0ll;
  REP(i, N) {
    // debug(a[i]);
    // debug(b[i]);
    ll tmp = b[i] - a[i];
    if (tmp < 0) cost += tmp;
    else {
      if (tmp <= 1) continue;
      cost += tmp / 2;
    }
    // debug(i);
    // debug(cost);
  }

  bool ok = true;
  if (cost < 0) ok = false;

  cout << (ok ? "Yes" : "No") << endl;
  
  return 0;
}