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 main() {
  long long n;
  cin >> n;
  long long a[n], b[n], ans[n], f[n], s[n + 1];
  for (long long i = 0; i < (n); i++) {
    cin >> a[i];
    ans[i] = 0;
    s[i] = 0;
    f[i] = 0;
  }
  s[n] = 0;
  for (long long i = 0; i < (n); i++) cin >> b[i];
  vector<long long> p(n);
  p[0] = b[0];
  for (long long i = 1; i < n; i++) p[i] = p[i - 1] + b[i];
  long long c[n];
  long long temp = 0;
  for (long long i = 0; i < (n); i++) {
    c[i] = lower_bound(p.begin(), p.end(), a[i] + temp) - p.begin();
    temp += b[i];
  }
  for (long long i = 0; i < (n); i++) {
    long long x = a[i];
    if (c[i] > i) {
      if (i - 1 != -1)
        x -= (p[c[i] - 1] - p[i - 1]);
      else
        x -= p[c[i] - 1];
      ans[c[i]] += x;
      s[i] += 1;
      s[c[i]] -= 1;
    } else
      ans[i] += x;
  }
  f[0] = s[0];
  for (long long i = 1; i < n; i++) {
    f[i] = s[i] + f[i - 1];
  }
  for (long long i = 0; i < (n); i++) {
    ans[i] += f[i] * b[i];
    cout << ans[i] << " ";
  }
  cout << endl;
  return 0;
}
