#include <bits/stdc++.h>
using namespace std;
template <typename Q>
void inin(Q &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x = f ? -x : x;
}
int n, a[100010], b[100010];
long long sum[100010], ans[100010], tot[100010];
int erfen(int x, int l, int r) {
  int ll = l, rr = r, mid;
  while (ll < rr) {
    mid = (ll + rr) >> 1;
    if (sum[mid] - sum[l - 1] < x)
      ll = mid + 1;
    else
      rr = mid;
  }
  if (ll == n && sum[n] - sum[l - 1] < x) return n + 1;
  return ll;
}
int main() {
  inin(n);
  for (int i = (1); i <= (n); i++) inin(a[i]);
  for (int i = (1); i <= (n); i++) inin(b[i]), sum[i] = sum[i - 1] + b[i];
  for (int i = (1); i <= (n); i++) {
    int x = erfen(a[i], i, n);
    tot[x]--;
    ans[x] += a[i] - (sum[x - 1] - sum[i - 1]);
    tot[i] += tot[i - 1] + 1;
    ans[i] += 1LL * b[i] * tot[i];
  }
  for (int i = (1); i <= (n); i++) printf("%lld ", ans[i]);
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
template <typename Q>
void inin(Q &x) {
  x = 0;
  int f = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = 1;
    ch = getchar();
  }
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  x = f ? -x : x;
}
int n, a[100010], b[100010];
long long sum[100010], ans[100010], tot[100010];
int erfen(int x, int l, int r) {
  int ll = l, rr = r, mid;
  while (ll < rr) {
    mid = (ll + rr) >> 1;
    if (sum[mid] - sum[l - 1] < x)
      ll = mid + 1;
    else
      rr = mid;
  }
  if (ll == n && sum[n] - sum[l - 1] < x) return n + 1;
  return ll;
}
int main() {
  inin(n);
  for (int i = (1); i <= (n); i++) inin(a[i]);
  for (int i = (1); i <= (n); i++) inin(b[i]), sum[i] = sum[i - 1] + b[i];
  for (int i = (1); i <= (n); i++) {
    int x = erfen(a[i], i, n);
    tot[x]--;
    ans[x] += a[i] - (sum[x - 1] - sum[i - 1]);
    tot[i] += tot[i - 1] + 1;
    ans[i] += 1LL * b[i] * tot[i];
  }
  for (int i = (1); i <= (n); i++) printf("%lld ", ans[i]);
  return 0;
}

```