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() {
  int n;
  scanf("%d", &n);
  vector<long long> v(n + 1), t(n + 1, 0), cumT(n + 1), ans(n + 1, 0),
      cntNeg(n + 2, 0);
  for (int i = 1; i <= n; ++i) scanf("%lld", &v[i]);
  for (int i = 1; i <= n; ++i)
    scanf("%lld", &t[i]), cumT[i] += cumT[i - 1] + t[i];
  for (int i = 1; i <= n; ++i) {
    int l = i, r = n, mark = -1;
    while (l <= r) {
      int mid = (l + r) / 2;
      if (cumT[mid] - cumT[i - 1] >= v[i]) {
        mark = mid;
        r = mid - 1;
      } else {
        l = mid + 1;
      }
    }
    if (mark != -1) {
      ans[mark] += v[i] - cumT[mark] + cumT[i - 1];
      cntNeg[mark + 1]++;
    }
  }
  for (int i = 1; i <= n; ++i) cntNeg[i] += cntNeg[i - 1];
  for (int i = 1; i <= n; ++i) {
    printf("%lld ", 1LL * t[i] * (i - cntNeg[i]) + ans[i]);
  }
  return 0;
}
