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 (GCC 5.4.1) (judge verdict: Accepted)

#include <stdio.h>

int main(void) {

  int n;
  scanf("%d", &n);
  long a[n],b[n];
  for (int i = 0; i < n; i++) {
    scanf("%ld", &a[i]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%ld", &b[i]);
  }
  long sum_a = 0;
  long sum_b = 0;
  for (int i = 0; i < n; i++) {
    sum_a += a[i];
    sum_b += b[i];
  }
  if (sum_a > sum_b) {
    printf("No\n");
    return 0;
  }
  long operation = sum_b-sum_a;
  long stock = 0;
  long add_to_a;
  for (int i = 0; i < n; i++) {
    if (a[i] > b[i]) {
      if (a[i]-b[i] > operation) {
        printf("No\n");
        return 0;
      }
      operation -= a[i]-b[i];
      stock += (a[i]-b[i])*2;
      b[i] = a[i];
    }
  }
  for (int i = 0; i < n; i++) {
    if (b[i] > a[i]) {
      if ((b[i]-a[i])/2 <= stock/2) {
        add_to_a = (b[i]-a[i])/2*2;
        a[i] += add_to_a;
        stock -= add_to_a;
      } else {
        a[i] += stock;
        stock = 0;
      }
      if (b[i] > a[i]) {
        if (b[i]-a[i] > operation) {
          printf("No\n");
          return 0;
        }
        operation -= b[i]-a[i];
      }
    }
  }
  printf("Yes\n");

  return 0;
}