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;
int dps[1022], pk;
int solve(string s) {
  for (int i = 0; i < s.size(); i++) {
    dps[i] = 1;
  }
  for (int i = 1; i < s.size(); i++) {
    int maxi = 0;
    for (int j = i - 1; j >= 0; j--) {
      if (s[i] == s[j]) {
        int t = maxi;
        maxi = max(maxi, dps[j]);
        dps[j] = max(dps[j], t + 2);
      } else {
        maxi = max(maxi, dps[j]);
      }
    }
  }
  int ans = 0;
  for (int i = 0; i < s.size(); i++) ans = max(ans, dps[i]);
  return ans;
}
string lltos(int t) {
  stringstream ss;
  ss << t;
  string b;
  ss >> b;
  return b;
}
long long stoll(string t) {
  stringstream ss;
  ss << t;
  long long b;
  ss >> b;
  return b;
}
int check1(int a, int b, int c) {
  int l = min(max(a, b) - min(a, b), c);
  int ans = min(a, b) + l;
  c -= l;
  ans += c / 2;
  return ans * 2;
}
int check2(int a, int b, int c) {
  int ans;
  for (int i = 0; i <= c; i++) {
    int x = a + i;
    int y = b + (c - i);
    ans = max(ans, min(x, y) * 2);
  }
  return ans;
}
long long modulo_Euclidian(long long a, long long b) {
  long long m = a % b;
  if (m < 0) {
    m = (b < 0) ? m - b : m + b;
  }
  return m;
}
const int MOD = 1e9 + 7, N = 1e5 + 4;
int n;
long long a[N], dop[N], kol[N], b[N], t[N];
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i <= n; i++) {
    cin >> t[i];
    b[i] = b[i - 1] + t[i];
  }
  for (int i = 1; i <= n; i++) {
    int pos = upper_bound(b + i, b + 1 + n, a[i] + b[i - 1]) - b - 1;
    kol[pos]++;
    kol[i - 1]--;
    long long r = a[i] - (b[pos] - b[i - 1]);
    dop[pos + 1] += r;
  }
  for (int i = n; i >= 1; i--) {
    kol[i] += kol[i + 1];
  }
  for (int i = 1; i <= n; i++) {
    long long ans = kol[i] * t[i] + dop[i];
    cout << ans << ' ';
  }
  return 0;
}
