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;
int n, l, r, id;
long long a[100005], b[100005] = {0}, f[100005] = {0}, g[100005] = {0};
int main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) cin >> b[i], b[i] += b[i - 1];
  for (int i = 1; i <= n; i++) {
    l = i, r = n, id = -1;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (b[mid] - b[i - 1] <= a[i])
        id = mid, l = mid + 1;
      else
        r = mid - 1;
    }
    if (id != -1) {
      f[i]++;
      f[id + 1]--;
      g[id + 1] += a[i] - (b[id] - b[i - 1]);
    } else {
      g[i] += a[i];
    }
  }
  for (int i = 1; i <= n; i++) f[i] += f[i - 1];
  for (int i = 1; i <= n; i++)
    f[i] = f[i] * (b[i] - b[i - 1]) + g[i], cout << f[i] << " ";
}
