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;
int tab[100005];
int pom[100005];
long long tem[100005];
long long odp[100005];
const int stala = (1 << 17);
int tr[2 * stala + 5];
void up(int x, int y) {
  x += stala, y += stala;
  while (x < y) {
    if (x % 2 == 1) {
      tr[x]++;
      x++;
    }
    if (y % 2 == 0) {
      tr[y]++;
      y--;
    }
    x /= 2;
    y /= 2;
  }
  if (x == y) tr[x]++;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(NULL);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> tab[i];
  }
  for (int i = 1; i <= n; i++) {
    cin >> tem[i];
    pom[i] = tem[i];
  }
  for (int i = n; i > 0; i--) {
    tem[i] += tem[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    if (tab[i] <= pom[i]) {
      odp[i] += tab[i];
      continue;
    }
    int l = i, p = n + 1;
    while (l + 1 < p) {
      int s = (l + p) / 2;
      if (tem[i] - tem[s + 1] > tab[i]) {
        p = s;
      } else {
        l = s;
      }
    }
    up(i, l);
    int sum = tem[i] - tem[l + 1];
    odp[l + 1] += tab[i] - sum;
  }
  for (int i = 1; i <= n; i++) {
    int x = i + stala;
    int ile = 0;
    while (x > 0) {
      ile += tr[x];
      x /= 2;
    }
    long long sum = (long long)pom[i] * ile + odp[i];
    cout << sum << " ";
  }
  return 0;
}
