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>
long long int MOD = 1000000007;
using namespace std;
template <class T>
inline void sary(T* st, T* nd) {
  while (st < nd) cin >> *st++;
}
template <class T>
inline double sarys(T* st, T* nd, T sum) {
  while (st < nd) {
    cin >> *st++;
    sum += *st;
  }
  return sum;
}
template <class T>
inline void saryp(T* st, T* nd) {
  while (st < nd) {
    cout << *st++ << " ";
  }
  cout << endl;
}
const int N = 1e6 + 5;
int main() {
  int n;
  cin >> n;
  long long int update[n + 10];
  long long int adhoora[n + 10];
  for (int i = 1; i <= n; i++) {
    update[i] = 0;
  }
  for (int i = 1; i <= n; i++) {
    adhoora[i] = 0;
  }
  long long int arr[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> arr[i];
  }
  long long int days[n + 1];
  for (int i = 1; i <= n; i++) {
    cin >> days[i];
  }
  long long int prefix[n + 1];
  prefix[0] = 0;
  for (int i = 1; i <= n; i++) {
    prefix[i] = prefix[i - 1] + days[i];
  }
  for (int i = 1; i <= n; i++) {
    long long int minus = prefix[i - 1];
    int low = i;
    int high = n;
    int ans = -1;
    while (low <= high) {
      int mid = low + (high - low) / 2;
      if (prefix[mid] - minus <= arr[i]) {
        ans = mid;
        low = mid + 1;
      } else {
        high = mid - 1;
      }
    }
    if (ans == -1) {
      adhoora[i] += arr[i];
    } else {
      update[i] += 1 * 1ll;
      update[ans + 1] += -1 * 1ll;
      if (arr[i] > prefix[ans] - minus) {
        long long int value_ad = arr[i] - (prefix[ans] - minus);
        adhoora[ans + 1] += value_ad;
      }
    }
  }
  update[0] = 0 * 1ll;
  for (int i = 1; i <= n; i++) {
    update[i] = update[i - 1] + update[i];
  }
  for (int i = 1; i <= n; i++) {
    cout << update[i] * 1ll * days[i] * 1ll + adhoora[i] * 1ll << " ";
  }
  cout << "\n";
  return 0;
}
