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;
long long t[100005];
long long sum[100005];
struct hs {
  long long v;
  int t;
  int x;
  bool s;
} v[100005 * 2];
long long sp[100005];
int get(int x) {
  int l = x, r = n;
  if (sum[r] - sum[l - 1] < v[x * 2 - 1].v) return n + 1;
  while (l < r) {
    int mid = (l + r) / 2;
    if (sum[mid] - sum[x - 1] < v[x * 2 - 1].v)
      l = mid + 1;
    else
      r = mid;
  }
  sp[x] = sum[l] - sum[x - 1] - v[x * 2 - 1].v;
  return l;
}
bool cmp(hs a, hs b) { return a.t < b.t; }
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= 2 * n; i += 2) {
    cin >> v[i].v;
    v[i + 1].v = v[i].v;
    v[i].s = 0;
    v[i + 1].s = 1;
    v[i].t = i / 2 + 1;
    v[i].x = v[i + 1].x = i / 2 + 1;
  }
  sum[0] = 0;
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
    sum[i] = sum[i - 1] + t[i];
  }
  for (int i = 1; i <= n; i++) {
    v[i * 2].t = get(i);
  }
  sort(v + 1, v + 2 * n + 1, cmp);
  int tim = 1;
  int cnt = 0;
  for (int i = 1; i <= n; i++) {
    long long ans = 0;
    while (v[tim].t == i) {
      if (v[tim].s == 0) {
        cnt++;
      } else {
        ans += (t[i] - sp[v[tim].x]);
        cnt--;
      }
      tim++;
    }
    ans = ans + cnt * t[i];
    cout << ans << ' ';
  }
  return 0;
}
