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;
const int N = 111111;
const long long INF = (long long)1e17;
int n;
long long u[N], v[N], a[4 * N], t[4 * N];
void init(int root, int rl, int rr) {
  if (rl == rr) {
    t[root] = INF;
  } else {
    int rm = (rl + rr) / 2;
    init(root * 2 + 1, rl, rm);
    init(root * 2 + 2, rm + 1, rr);
    t[root] = min(t[root * 2 + 1], t[root * 2 + 2]);
  }
}
void down(int root) {
  if (a[root] != 0) {
    a[root * 2 + 1] += a[root];
    a[root * 2 + 2] += a[root];
    t[root * 2 + 1] += a[root];
    t[root * 2 + 2] += a[root];
    a[root] = 0;
  }
}
void update(int root, int rl, int rr, int l, int r, long long val) {
  if (l > r) {
    return;
  }
  if (l <= rl && rr <= r) {
    a[root] += val;
    t[root] += val;
    return;
  }
  int rm = (rl + rr) / 2;
  down(root);
  update(root * 2 + 1, rl, rm, l, min(r, rm), val);
  update(root * 2 + 2, rm + 1, rr, max(l, rm + 1), r, val);
  t[root] = min(t[root * 2 + 1], t[root * 2 + 2]);
}
pair<int, long long> query(int root, int rl, int rr) {
  if (rl == rr) {
    return make_pair(rl, t[root]);
  }
  int rm = (rl + rr) / 2;
  down(root);
  t[root] = min(t[root * 2 + 1], t[root * 2 + 2]);
  if (t[root * 2 + 1] <= t[root * 2 + 2]) {
    return query(root * 2 + 1, rl, rm);
  } else {
    return query(root * 2 + 2, rm + 1, rr);
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) {
    scanf("%lld", &u[i]);
  }
  for (int i = 0; i < n; i++) {
    scanf("%lld", &v[i]);
  }
  init(0, 0, n - 1);
  for (int i = 0, c = 0; i < n; i++) {
    update(0, 0, n - 1, i, i, -INF + u[i]);
    c++;
    long long ans = 0;
    while (true) {
      pair<int, long long> tmp = query(0, 0, n - 1);
      if (tmp.second > v[i]) {
        break;
      }
      c--;
      ans += tmp.second;
      update(0, 0, n - 1, tmp.first, tmp.first, INF - tmp.second);
    }
    ans += c * v[i];
    update(0, 0, n - 1, 0, i, -v[i]);
    printf("%lld ", ans);
  }
  puts("");
  return 0;
}
