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(...) overload_rep(__VA_ARGS__, rep_1, rep_0) (__VA_ARGS__)
#define rep_0(i, end) for (int i = 0; i < (end); ++i)
#define rep_1(i, fr, ba) for (int i = (fr); i <= (ba); ++i)
#define overload_rep(arg_0, arg_1, arg_2, name, ...) name
#define all(box) box.begin(), box.end()
 
template <class T, class U> inline bool chmax(T& lhs, const U& rhs) { if (lhs < rhs) { lhs = rhs; return 1; } return 0; }
template <class T, class U> inline bool chmin(T& lhs, const U& rhs) { if (lhs > rhs) { lhs = rhs; return 1; } return 0; }

int main() {

  int N;
  std::cin >> N;
  std::vector<long long> a(N), b(N);
  rep (i, N) std::cin >> a[i];
  rep (i, N) std::cin >> b[i];

  long long one = 0, two = 0, steps = 0;
  rep (i, N) {
    steps += b[i] - a[i];
    if (a[i] < b[i]) two += (b[i] - a[i] + 1) / 2;
    else one += a[i] - b[i];
  }

  puts(one <= steps && two <= steps ? "Yes" : "No");
  return 0;

}