#include <bits/stdc++.h>
using namespace std;
long long int extra[100005], full[100005], sum[100005], t[100005], v[100005];
int main() {
  int n, i;
  cin >> n;
  for (i = 0; i < n; i++) full[i] = extra[i] = 0;
  for (i = 0; i < n; i++) cin >> v[i];
  for (i = 0; i < n; i++) cin >> t[i];
  sum[0] = t[0];
  for (i = 1; i < n; i++) sum[i] = sum[i - 1] + t[i];
  for (i = 0; i < n; i++) {
    long long int curr = 0;
    curr = sum[n - 1];
    if (i > 0) curr -= sum[i - 1];
    if (curr <= v[i]) {
      full[i] += 1;
      continue;
    }
    if (t[i] >= v[i]) {
      extra[i] += v[i];
      continue;
    }
    int lo = i, hi = n - 1, m;
    while (lo < hi - 1) {
      m = (lo + hi) / 2;
      curr = sum[m];
      if (i > 0) curr -= sum[i - 1];
      if (curr > v[i])
        hi = m;
      else
        lo = m;
    }
    curr = sum[lo];
    if (i > 0) curr -= sum[i];
    if (curr > v[i]) hi = lo;
    curr = sum[hi - 1];
    if (i > 0) curr -= sum[i - 1];
    full[i] += 1;
    full[hi] -= 1;
    extra[hi] += v[i] - curr;
  }
  for (i = 1; i < n; i++) full[i] += full[i - 1];
  for (i = 0; i < n; i++) cout << full[i] * t[i] + extra[i] << ' ';
  cout << '\n';
  return 0;
}

### Prompt

Develop a solution in CPP to the problem described below:

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 int extra[100005], full[100005], sum[100005], t[100005], v[100005];
int main() {
  int n, i;
  cin >> n;
  for (i = 0; i < n; i++) full[i] = extra[i] = 0;
  for (i = 0; i < n; i++) cin >> v[i];
  for (i = 0; i < n; i++) cin >> t[i];
  sum[0] = t[0];
  for (i = 1; i < n; i++) sum[i] = sum[i - 1] + t[i];
  for (i = 0; i < n; i++) {
    long long int curr = 0;
    curr = sum[n - 1];
    if (i > 0) curr -= sum[i - 1];
    if (curr <= v[i]) {
      full[i] += 1;
      continue;
    }
    if (t[i] >= v[i]) {
      extra[i] += v[i];
      continue;
    }
    int lo = i, hi = n - 1, m;
    while (lo < hi - 1) {
      m = (lo + hi) / 2;
      curr = sum[m];
      if (i > 0) curr -= sum[i - 1];
      if (curr > v[i])
        hi = m;
      else
        lo = m;
    }
    curr = sum[lo];
    if (i > 0) curr -= sum[i];
    if (curr > v[i]) hi = lo;
    curr = sum[hi - 1];
    if (i > 0) curr -= sum[i - 1];
    full[i] += 1;
    full[hi] -= 1;
    extra[hi] += v[i] - curr;
  }
  for (i = 1; i < n; i++) full[i] += full[i - 1];
  for (i = 0; i < n; i++) cout << full[i] * t[i] + extra[i] << ' ';
  cout << '\n';
  return 0;
}

```