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;
const long long MAX = 1e5 + 5, inf = 1e17;
long long arr[4 * MAX] = {};
long long tree[4 * MAX] = {};
long long lazy[4 * MAX] = {};
void build_tree(long long node, long long a, long long b) {
  if (a > b) return;
  if (a == b) {
    tree[node] = arr[a];
    return;
  }
  build_tree(node * 2, a, (a + b) / 2);
  build_tree(node * 2 + 1, 1 + (a + b) / 2, b);
  tree[node] = tree[node * 2] + tree[node * 2 + 1];
}
void update_tree(long long node, long long a, long long b, long long i,
                 long long j, long long value) {
  if (lazy[node] != 0) {
    tree[node] += lazy[node];
    if (a != b) {
      lazy[node * 2] += lazy[node];
      lazy[node * 2 + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (a > b || a > j || b < i) return;
  if (a >= i && b <= j) {
    tree[node] += value;
    if (a != b) {
      lazy[node * 2] += value;
      lazy[node * 2 + 1] += value;
    }
    return;
  }
  update_tree(node * 2, a, (a + b) / 2, i, j, value);
  update_tree(1 + node * 2, 1 + (a + b) / 2, b, i, j, value);
  tree[node] = tree[node * 2] + tree[node * 2 + 1];
}
long long query_tree(long long node, long long a, long long b, long long i,
                     long long j) {
  if (a > b || a > j || b < i) return 0;
  if (lazy[node] != 0) {
    tree[node] += lazy[node];
    if (a != b) {
      lazy[node * 2] += lazy[node];
      lazy[node * 2 + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (a >= i && b <= j) return tree[node];
  long long q1 = query_tree(node * 2, a, (a + b) / 2, i, j);
  long long q2 = query_tree(1 + node * 2, 1 + (a + b) / 2, b, i, j);
  long long res = q1 + q2;
  return res;
}
int main() {
  long long n;
  cin >> n;
  long long el[n], tmp[n];
  for (long long i = 0; i < n; i++) cin >> el[i];
  for (long long i = 0; i < n; i++) cin >> tmp[i];
  vector<long long> pre(n + 1);
  for (long long i = 0; i < n; i++)
    pre[i] = (i == 0) ? tmp[i] : tmp[i] + pre[i - 1];
  for (long long i = 0; i < n; i++) arr[i] = 0;
  build_tree(1, 0, n - 1);
  memset(lazy, 0, sizeof lazy);
  pre[n] = inf;
  long long ans[n + 5];
  for (int i = 0; i < n + 5; i++) ans[i] = 0;
  for (long long i = 0; i < n; i++) {
    long long e = el[i];
    if (i) {
      e += pre[i - 1];
    }
    vector<long long>::iterator it = lower_bound(pre.begin(), pre.end(), e);
    if (*it > e && (it - pre.begin() == 0)) {
      ans[0] += el[i];
      continue;
    }
    if (*it > e) it--;
    long long idx = it - pre.begin();
    long long rem = e - pre[idx];
    ans[idx + 1] += rem;
    update_tree(1, 0, n - 1, i, idx, 1);
  }
  for (int i = 0; i < n; i++) {
    long long res = tmp[i] * query_tree(1, 0, n - 1, i, i);
    res += ans[i];
    cout << res << " ";
  }
  return 0;
}
