#include <bits/stdc++.h>
using namespace std;
long long ser(long long in, std::vector<long long> v, long long size,
              long long it) {
  for (long long i = in; i < size; i++) {
    if (v[i] > it) return i - 1;
  }
  return size - 1;
}
int main() {
  long long n;
  cin >> n;
  long long a[n];
  long long b[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
  }
  std::vector<long long> fre(n + 1, 0);
  for (long long i = 1; i <= n; i++) {
    fre[i] = fre[i - 1] + b[i - 1];
  }
  vector<long long> mul(n + 2, 0);
  vector<long long> addd(n + 2, 0);
  for (long long i = 1; i <= n; i++) {
    long long temp =
        lower_bound(fre.begin(), fre.end(), a[i - 1] + fre[i - 1]) -
        fre.begin();
    temp--;
    mul[i] += 1;
    mul[temp + 1] -= 1;
    addd[temp + 1] += (max(0ll, a[i - 1] + fre[i - 1] - fre[temp]));
  }
  for (long long i = 0; i <= n; i++) {
    mul[i + 1] += mul[i];
  }
  for (long long i = 1; i <= n; i++) cout << mul[i] * b[i - 1] + addd[i] << " ";
}

### Prompt

Please formulate a Cpp solution to the following problem:

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.

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long ser(long long in, std::vector<long long> v, long long size,
              long long it) {
  for (long long i = in; i < size; i++) {
    if (v[i] > it) return i - 1;
  }
  return size - 1;
}
int main() {
  long long n;
  cin >> n;
  long long a[n];
  long long b[n];
  for (long long i = 0; i < n; i++) {
    cin >> a[i];
  }
  for (long long i = 0; i < n; i++) {
    cin >> b[i];
  }
  std::vector<long long> fre(n + 1, 0);
  for (long long i = 1; i <= n; i++) {
    fre[i] = fre[i - 1] + b[i - 1];
  }
  vector<long long> mul(n + 2, 0);
  vector<long long> addd(n + 2, 0);
  for (long long i = 1; i <= n; i++) {
    long long temp =
        lower_bound(fre.begin(), fre.end(), a[i - 1] + fre[i - 1]) -
        fre.begin();
    temp--;
    mul[i] += 1;
    mul[temp + 1] -= 1;
    addd[temp + 1] += (max(0ll, a[i - 1] + fre[i - 1] - fre[temp]));
  }
  for (long long i = 0; i <= n; i++) {
    mul[i + 1] += mul[i];
  }
  for (long long i = 1; i <= n; i++) cout << mul[i] * b[i - 1] + addd[i] << " ";
}

```