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;
long long binarySearch(long long k, long long low, long long high,
                       long long A[], long long sub) {
  if (low <= high) {
    long long mid = low + (high - low) / 2;
    if (A[mid] - sub < (k)) {
      return binarySearch(k, mid + 1, high, A, sub);
    } else if (A[mid] - sub >= (k)) {
      if (mid > low) {
        if (A[mid - 1] - sub >= (k))
          return binarySearch(k, low, mid - 1, A, sub);
        return mid;
      }
      return mid;
    }
  }
  return -1;
}
void update(long long A[], long long sum[], long long n, long long val,
            long long pre[], long long id, long long maxI) {
  if (maxI == -1) {
    A[id] += 1;
  } else {
    A[id] += 1;
    A[maxI] -= 1;
    if (maxI - 1 >= 0) {
      if (id >= 1)
        sum[maxI] += val - (pre[maxI - 1] - pre[id - 1]);
      else
        sum[maxI] += val - (pre[maxI - 1]);
    } else
      sum[maxI] += val;
  }
}
int main() {
  long long n;
  cin >> n;
  long long v[n];
  for (long long i = 0; i < n; i++) {
    cin >> v[i];
  }
  long long t[n];
  for (long long i = 0; i < n; i++) {
    cin >> t[i];
  }
  long long pre[n];
  pre[0] = t[0];
  for (long long i = 1; i < n; i++) {
    pre[i] = t[i] + pre[i - 1];
  }
  long long A[n];
  long long sum[n];
  for (long long i = 0; i < n; i++) {
    A[i] = sum[i] = 0;
  }
  for (long long i = 0; i < n; i++) {
    long long maxI;
    if (i == 0) {
      maxI = binarySearch(v[i], i, n - 1, pre, 0);
    } else
      maxI = binarySearch(v[i], i, n - 1, pre, pre[i - 1]);
    update(A, sum, n, v[i], pre, i, maxI);
  }
  for (long long i = 1; i < n; i++) {
    A[i] = A[i - 1] + A[i];
  }
  for (long long i = 0; i < n; i++) {
    sum[i] = A[i] * t[i] + sum[i];
  }
  for (long long i = 0; i < n; i++) {
    cout << sum[i] << " ";
  }
  cout << endl;
}
