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;
inline int read() {
  int x;
  char c;
  int f = 1;
  while ((c = getchar()) != '-' && (c > '9' || c < '0'))
    ;
  if (c == '-') c = getchar(), f = -1;
  x = c ^ '0';
  while ((c = getchar()) >= '0' && c <= '9')
    x = (x << 1) + (x << 3) + (c ^ '0');
  return x * f;
}
inline long long readll() {
  long long x;
  char c;
  int f = 1;
  while ((c = getchar()) != '-' && (c > '9' || c < '0'))
    ;
  if (c == '-') c = getchar(), f = -1;
  x = c ^ '0';
  while ((c = getchar()) >= '0' && c <= '9')
    x = (x << 1ll) + (x << 3ll) + (c ^ '0');
  return x * f;
}
const int maxn = 1e5 + 10;
long long a[maxn];
long long b[maxn], c[maxn];
long long ans[maxn];
int main() {
  int n = read();
  for (register int i = 1, iend = n; i <= iend; ++i) a[i] = read();
  for (register int i = 1, iend = n; i <= iend; ++i) {
    b[i] = read();
    b[i] += b[i - 1];
  }
  for (register int i = 1, iend = n; i <= iend; ++i) {
    long long u = a[i] + b[i - 1];
    int x = lower_bound(b + 1, b + n + 1, u) - b;
    c[i] += 1, c[x] -= 1;
    ans[x] += u - b[x - 1];
  }
  for (register int i = 1, iend = n; i <= iend; ++i) {
    c[i] += c[i - 1];
    ans[i] += c[i] * (b[i] - b[i - 1]);
    printf("%lld%c", ans[i], i == iend ? '\n' : ' ');
  }
  return 0;
}
