Alice likes snow a lot! Unfortunately, this year's winter is already over, and she can't expect to have any more of it. Bob has thus bought her a gift — a large snow maker. He plans to make some amount of snow every day. On day i he will make a pile of snow of volume Vi and put it in her garden.

Each day, every pile will shrink a little due to melting. More precisely, when the temperature on a given day is Ti, each pile will reduce its volume by Ti. If this would reduce the volume of a pile to or below zero, it disappears forever. All snow piles are independent of each other. 

Note that the pile made on day i already loses part of its volume on the same day. In an extreme case, this may mean that there are no piles left at the end of a particular day.

You are given the initial pile sizes and the temperature on each day. Determine the total volume of snow melted on each day. 

Input

The first line contains a single integer N (1 ≤ N ≤ 105) — the number of days. 

The second line contains N integers V1, V2, ..., VN (0 ≤ Vi ≤ 109), where Vi is the initial size of a snow pile made on the day i.

The third line contains N integers T1, T2, ..., TN (0 ≤ Ti ≤ 109), where Ti is the temperature on the day i.

Output

Output a single line with N integers, where the i-th integer represents the total volume of snow melted on day i.

Examples

Input

3
10 10 5
5 7 2


Output

5 12 4


Input

5
30 25 20 15 10
9 10 12 4 13


Output

9 20 35 11 25

Note

In the first sample, Bob first makes a snow pile of volume 10, which melts to the size of 5 on the same day. On the second day, he makes another pile of size 10. Since it is a bit warmer than the day before, the first pile disappears completely while the second pile shrinks to 3. At the end of the second day, he has only a single pile of size 3. On the third day he makes a smaller pile than usual, but as the temperature dropped too, both piles survive till the end of the day.
#include <bits/stdc++.h>
using namespace std;
long long int n, V[100002], T[100002];
long long int Tsum[100002];
long long int diff[100002], counts[100002];
long long int ceilSearch(long long int arr[], int low, int high,
                         long long int x) {
  int mid;
  if (x <= arr[low]) return low;
  if (x > arr[high]) return -1;
  mid = (low + high) / 2;
  if (arr[mid] == x)
    return mid;
  else if (arr[mid] < x) {
    if (mid + 1 <= high && x <= arr[mid + 1])
      return mid + 1;
    else
      return ceilSearch(arr, mid + 1, high, x);
  } else {
    if (mid - 1 >= low && x > arr[mid - 1])
      return mid;
    else
      return ceilSearch(arr, low, mid - 1, x);
  }
}
int main() {
  cin >> n;
  int i;
  for (i = 0; i < n; i++) {
    cin >> V[i];
  }
  for (i = 0; i < n; i++) {
    cin >> T[i];
  }
  Tsum[0] = T[0];
  for (i = 1; i < n; i++) {
    Tsum[i] = Tsum[i - 1] + T[i];
  }
  int index;
  for (i = 0; i < n; i++) {
    if (i == 0) {
      index = ceilSearch(Tsum, 0, n - 1, V[i]);
    } else {
      index = ceilSearch(Tsum, 0, n - 1, V[i] + Tsum[i - 1]);
    }
    if (index == -1) {
      counts[i] += 1;
    } else if (index == i) {
      diff[index] += V[i];
    } else {
      counts[i] += 1;
      counts[index] -= 1;
      diff[index] +=
          (T[index] - ((Tsum[index]) - (V[i] + (i > 0 ? Tsum[i - 1] : 0))));
    }
  }
  int ctr = 0;
  for (i = 0; i < n; i++) {
    ctr += counts[i];
    cout << ctr * T[i] + diff[i] << " ";
  }
  return 0;
}
