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;
struct inter {
  long long int s, e;
};
int main() {
  long long n;
  cin >> n;
  long long V[n], T[n], Rsum[n], Lsum[n];
  for (long long int i = 0; i < n; ++i) cin >> V[i];
  long long sum = 0;
  for (long long int i = 0; i < n; ++i) {
    cin >> T[i];
    sum += T[i];
    Lsum[i] = sum;
  }
  sum = 0;
  for (long long int i = n - 1; i >= 0; --i) {
    sum += T[i];
    Rsum[i] = sum;
  }
  vector<inter> AN;
  long long answer[n];
  vector<long long> LsumA(Lsum, Lsum + n);
  long long int interval[n];
  for (int i = 0; i < n; ++i) {
    answer[i] = 0;
    interval[i] = 0;
  }
  for (int i = 0; i < n; i++) {
    struct inter p;
    if (V[i] >= Rsum[i]) {
      p.s = i;
      p.e = n;
      AN.push_back(p);
    } else {
      if (V[i] <= T[i]) {
        answer[i] += V[i];
      } else {
        p.s = i;
        p.e = i;
        long long tf = Lsum[i] - T[i] + V[i];
        long long ind =
            upper_bound(LsumA.begin(), LsumA.end(), tf) - LsumA.begin();
        if (ind >= n) {
          p.e = n;
        } else if (Lsum[ind - 1] == tf) {
          p.e = ind - 1;
        } else {
          p.e = max(ind - 1, p.e);
          answer[ind] += V[i] - (Lsum[ind - 1] - Lsum[i] + T[i]);
        }
        AN.push_back(p);
      }
    }
  }
  for (int i = 0; i < AN.size(); ++i) {
    if (AN[i].s < n) {
      interval[AN[i].s]++;
      if (AN[i].e + 1 < n) interval[AN[i].e + 1]--;
    }
  }
  for (int i = 0; i < n; ++i) {
    if (i) interval[i] += interval[i - 1];
    answer[i] += interval[i] * T[i];
    cout << answer[i] << " ";
  }
  return 0;
}
