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 double eps = 1e-8;
const int MAXN = 1e5 + 5;
const long long MOD = 1e9 + 7;
int n;
long long v[MAXN], t[MAXN];
long long qz[MAXN];
long long ew[MAXN];
long long bit[MAXN];
int lowbit(int x) { return x & (-x); }
void add(int x, long long y) {
  while (x <= n) {
    bit[x] += y;
    x += lowbit(x);
  }
}
long long sum(int x) {
  long long res = 0;
  while (x) {
    res += bit[x];
    x -= lowbit(x);
  }
  return res;
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%lld", &v[i]);
  for (int i = 1; i <= n; i++) {
    scanf("%lld", &t[i]);
    qz[i] = qz[i - 1] + t[i];
  }
  for (int i = 1; i <= n; i++) {
    int l = i;
    int r = n;
    int ind = n + 1;
    while (l <= r) {
      int mid = (r + l) / 2;
      if (qz[mid] - qz[i - 1] >= v[i]) {
        r = mid - 1;
        ind = mid;
      } else
        l = mid + 1;
    }
    if (ind == n + 1) {
      add(i, 1);
      add(n + 1, -1);
    } else {
      add(i, 1);
      add(ind, -1);
      ew[ind] += v[i] - (qz[ind - 1] - qz[i - 1]);
    }
  }
  for (int i = 1; i <= n; i++)
    printf("%lld%c", sum(i) * t[i] + ew[i], i == n ? '\n' : ' ');
  return 0;
}
