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 int N = 400005;
int tree[N], lazy[N];
void updateRange(long long node, long long start, long long end, long long l,
                 long long r, long long val) {
  if (lazy[node] != 0) {
    tree[node] += (end - start + 1) * lazy[node];
    if (start != end) {
      lazy[node * 2] += lazy[node];
      lazy[node * 2 + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (start > end or start > r or end < l) return;
  if (start >= l and end <= r) {
    tree[node] += (end - start + 1) * val;
    if (start != end) {
      lazy[node * 2] += val;
      lazy[node * 2 + 1] += val;
    }
    return;
  }
  long long mid = (start + end) / 2;
  updateRange(node * 2, start, mid, l, r, val);
  updateRange(node * 2 + 1, mid + 1, end, l, r, val);
  tree[node] = tree[node * 2] + tree[node * 2 + 1];
}
long long queryRange(long long node, long long start, long long end,
                     long long l, long long r) {
  if (start > end or start > r or end < l) return 0;
  if (lazy[node] != 0) {
    tree[node] += (end - start + 1) * lazy[node];
    if (start != end) {
      lazy[node * 2] += lazy[node];
      lazy[node * 2 + 1] += lazy[node];
    }
    lazy[node] = 0;
  }
  if (start >= l and end <= r) return tree[node];
  long long mid = (start + end) / 2;
  long long p1 = queryRange(node * 2, start, mid, l, r);
  long long p2 = queryRange(node * 2 + 1, mid + 1, end, l, r);
  return (p1 + p2);
}
long long n;
long long V[100005], T[100005], S[100005];
long long answer[100005];
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(nullptr);
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> V[i];
  for (long long i = 1; i <= n; i++) cin >> T[i], S[i] = S[i - 1] + T[i];
  for (long long i = 1; i <= n; i++) {
    long long lo = i, hi = n, ans = i - 1, mid;
    while (lo <= hi) {
      mid = (lo + hi) / 2;
      long long melted = S[mid] - S[i - 1];
      if (melted <= V[i]) {
        ans = mid;
        lo = mid + 1;
      } else {
        hi = mid - 1;
      }
    }
    updateRange(1, 1, n, i, ans, 1);
    long long melted = S[ans] - S[i - 1];
    long long remain = V[i] - melted;
    answer[ans + 1] += remain;
  }
  for (int i = 1; i <= n; i++) {
    long long bing = queryRange(1, 1, n, i, i);
    answer[i] += bing * T[i];
    cout << answer[i] << " ";
  }
}
