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.
import java.util.Scanner;

public class App {
	public static int binarySearch(long arr[], int l, int r, long x) {
		int m;
		while (l < r) {
			m = l + (r - l) / 2;
			if (x < arr[m])
				r = m;
			else
				l = m + 1;
		}
		if (arr[arr.length - 1] < x) {
			r = -1;
		}
		return r;
	}

	public static void main(String[] args) {
		Scanner inp = new Scanner(System.in);
		int n = inp.nextInt();
		long[] v = new long[n];
		long[] t = new long[n];
		for (int i = 0; i < n; i++) {
			v[i] = inp.nextLong();
		}
		for (int i = 0; i < n; i++) {
			t[i] = inp.nextLong();
		}
		for (int i = 1; i < n; i++) {
			t[i] += t[i - 1];
			v[i] += t[i - 1];
		}
		long[] op = new long[n];
		long[] lv = new long[n];
		int ld;
		ld = binarySearch(t, 0, n - 1, v[0]);
		op[0] += 1;
		if (ld != -1) {
			op[ld] -= 1;
			if (ld != 0) {
				lv[ld] += v[0] - t[ld - 1];
			} else {
				lv[ld] += v[0];
			}
		}
		for (int i = 1; i < n; i++) {
			ld = binarySearch(t, 0, n - 1, v[i]);
			op[i] += 1;
			if (ld != -1) {
				op[ld] -= 1;
				lv[ld] += v[i] - t[ld - 1];
			}
		}
		for (int i = 1; i < n; i++) {
			op[i] += op[i - 1];
		}
		op[0] *= t[0];
		op[0] += lv[0];
		for (int i = 1; i < n; i++) {
			op[i] *= (t[i] - t[i - 1]);
			op[i] += lv[i];
		}
		for (int i = 0; i < n; i++) {
			System.out.print(op[i] + " ");
		}
		System.out.println();
		inp.close();
	}

}
