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 <bits/stdc++.h>
using namespace std;

// #include <boost/algorithm/string/join.hpp>
// using boost::algorithm::join;
//
// #include "boost/multi_array.hpp"
// boost::multi_array<int, 3> dp;
// dp.resize(boost::extents[10][9][8]);
//
// #include <boost/multiprecision/cpp_dec_float.hpp>  // quite slowly
// #include <boost/multiprecision/cpp_int.hpp>
// namespace mp = boost::multiprecision; // mp::log10
// using bigint = boost::multiprecision::cpp_int;
// using rational = mp::cpp_rational;  // double d = static_cast<double>(r);

typedef long long ll;
typedef unsigned long long ull;
//typedef long double ld;

typedef map<int, int> mii;
typedef pair<int, int> pii;
typedef pair<double,double> pdd;
typedef set<int> si;
typedef set<ll> sll;
typedef vector<int> vi;
typedef vector<pair<int, int > > vii;
typedef vector<ll> vll;
typedef vector<double> vd;
typedef vector<vi> vvi;
typedef vector<vii> vvii;
typedef vector<vll> vvll;

//constexpr ll MOD = 1e9+7;
template<typename T>constexpr auto inf=numeric_limits<T>::max()/4;
constexpr int INF=inf<int>;
constexpr ll LINF=inf<ll>;

#define Sort(x) sort(begin(x),end(x))
#define Reverse(x) reverse(begin(x),end(x))
#define ABS(a,b) ((a)<(b)?(b)-(a):(a)-(b))
#define nbit(n) (1ull<<(n))
#define _ol3(_1,_2,_3,name,...)name
#define _rep(i,n)_repi(i,0,n)
#define _repi(i,a,b)for(int i=a,i##_l=(b);i<i##_l;++i)
#define REP(...) _ol3(__VA_ARGS__,_repi,_rep,)(__VA_ARGS__)
#define REPR(i,n) for(int i=n-1;i>=0;--i)
#define REPA(i,v) REP(i,(v).size())

// sort(begin(x), end(x), greater<ll>());
// lower_bound(a.begin(), a.end(), val): a[i] ≥ valである最左の位置
// upper_bound(a.begin(), a.end(), val): a[i] > valである最左の位置

// Returns minimum of a and b.
// If a is less b, a is set to b.
template<typename T>
T chmin(T& a, T b) {
  if (a > b) {
    a = b;
  }
  return a;
}

// Returns maximum of a and b.
// If a is less b, a is set to b.
template<typename T>
T chmax(T& a, T b) {
  if (a < b) {
    a = b;
  }
  return a;
}

template <typename T>
void dumpContents(const T& v, const string& msg)
{
  cerr << "### " << msg << " ###\n";
  for (const auto& x : v) {
    cerr << x << " ";
  }
  cerr << endl;
}

struct before_main_function {
  before_main_function() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    cout<<setprecision(20)<<fixed;
#define endl "\n"
  }
} before_main_function;

// ========== end of template ==========

int main(int argc, char** argv) {
  int N;
  cin >> N;
  vll A(N);
  vll B(N);

  for (auto& a : A) {
    cin >> a;
  }
  for (auto& b : B) {
    cin >> b;
  }

  ll sumA = accumulate(begin(A), end(A), 0);
  ll sumB = accumulate(begin(B), end(B), 0);

  ll sumX = sumB-sumA;
  if (sumX<0) {
    cout << "No" << endl;
    return 0;
  }

  ll ax = 0;
  ll bx = 0;

  REP(i, N) {
    ll a = A[i];
    ll b = B[i];
    if (a>=b) {
      bx += a-b;
    } else {
      // a < b;
      if ((b-a)%2==0) {
        ax += (b-a)/2;
      } else {
        ax += (b+1-a)/2;
        bx += 1;
      }
    }
  }

  if (sumX<ax || sumX<bx) {
    cout << "No" << endl;
    return 0;
  }

  if ((sumX-ax)*2 == (sumX-bx)*1) {
    cout << "Yes" << endl;
    return 0;
  }

  cout << "No" << endl;
  return 0;
}

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