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 = 1e5 + 123;
long long a[N], t[N], it[N * 6], lz[N * 6], ans[N], n, res[N];
void load() {}
void down(long long k, long long l, long long r) {
  it[k * 2] += lz[k];
  it[k * 2 + 1] += lz[k];
  lz[k * 2] += lz[k];
  lz[k * 2 + 1] += lz[k];
  lz[k] = 0;
}
void update(long long k, long long x, long long y, long long l, long long r,
            long long val) {
  if (x > r || y < l) return;
  if (x <= l && r <= y) {
    ++it[k];
    ++lz[k];
    return;
  }
  down(k, l, r);
  long long mid = (l + r) / 2;
  update(k * 2, x, y, l, mid, val);
  update(k * 2 + 1, x, y, mid + 1, r, val);
  return;
}
long long get(long long k, long long x, long long l, long long r) {
  if (x > r || x < l) return 0;
  if (x == l && x == r) return it[k];
  down(k, l, r);
  long long mid = (l + r) / 2;
  if (x >= l && x <= mid) return get(k * 2, x, l, mid);
  return get(k * 2 + 1, x, mid + 1, r);
}
void process() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  for (long long i = 1; i <= n; ++i) scanf("%lld", &t[i]);
  for (long long i = n; i > 0; --i) {
    res[i] = res[i + 1] + t[i];
    long long d = i, c = n;
    while (d < c) {
      long long mid = (d + c + 1) / 2;
      if (a[i] >= res[i] - res[mid + 1])
        d = mid;
      else
        c = mid - 1;
    }
    if (a[i] >= res[i] - res[d + 1]) {
      update(1, i, d, 1, n, 1);
      long long tmp = a[i] - (res[i] - res[d + 1]);
      ans[d + 1] += tmp;
    } else
      ans[i] += a[i];
  }
  for (long long i = 1; i <= n; ++i) {
    long long tmp = get(1, i, 1, n);
    printf("%lld ", tmp * t[i] + ans[i]);
  }
}
int main() {
  load();
  process();
  return 0;
}
