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 *V*<sub class="lower-index">*i*</sub> 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 *T*<sub class="lower-index">*i*</sub>, each pile will reduce its volume by *T*<sub class="lower-index">*i*</sub>. 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.
The first line contains a single integer *N* (1 ≤ *N* ≤ 10<sup class="upper-index">5</sup>) — the number of days. 

The second line contains *N* integers *V*<sub class="lower-index">1</sub>, *V*<sub class="lower-index">2</sub>, ..., *V*<sub class="lower-index">*N*</sub> (0 ≤ *V*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>), where *V*<sub class="lower-index">*i*</sub> is the initial size of a snow pile made on the day *i*.

The third line contains *N* integers *T*<sub class="lower-index">1</sub>, *T*<sub class="lower-index">2</sub>, ..., *T*<sub class="lower-index">*N*</sub> (0 ≤ *T*<sub class="lower-index">*i*</sub> ≤ 10<sup class="upper-index">9</sup>), where *T*<sub class="lower-index">*i*</sub> is the temperature on the day *i*.
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.
3
10 10 5
5 7 2
5 12 4
5
30 25 20 15 10
9 10 12 4 13
9 20 35 11 25
4
0 0 0 0
1 2 3 4
0 0 0 0
10
11 39 16 34 25 3 12 11 31 16
10 0 4 9 8 9 7 8 9 2
10 0 9 27 27 30 28 17 12 4
10
20 35 4 0 6 29 4 9 17 10
0 9 4 7 5 1 4 3 9 4
0 18 12 14 10 3 12 9 26 12
1
4
5
4
1
5
4
4
1
5
5
5
2
9 3
8 2
8 3
2
9 3
4 4
4 7
2
9 3
10 2
9 2
13
1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1
1 1 1 1 1 1 1 1 1 1 1 1 1
#!/usr/bin/env python3

import sys

n = int(sys.stdin.readline().strip())
vs = list(map(int, sys.stdin.readline().strip().split()))
ts = list(map(int, sys.stdin.readline().strip().split()))

# binsearch e in a
def bs(a, e, l, r):
	while r - l > 1:
		m = (l + r) // 2
		if a[m] > e:
			r = m
		else:
			l = m
	return l

# integral
its = [0 for _ in range(n + 1)]
for i, t in enumerate(ts):
	its[i + 1] = its[i] + t

# melt out days
mds = [bs(its, vs[i] + its[i], i, n + 1) for i in range(n)]

# result
res = [0 for _ in range(n)]

# add last day`s remains
for i in range(n):
	if mds[i] < n:
		res[mds[i]] += vs[i] + its[i] - its[mds[i]]

# O(n^2)
#for i in range(n):
#	for j in range(i, md[i]):
#		res[j] += ts[j]

# counter of right ends (a histogram)
rcounts = [0 for _ in range(n + 1)]
for md in mds:
	rcounts[md] += 1

# integral of the counter (except last count)
ircounts = [0 for _ in range(n + 1)]
for i in range (n):
	ircounts[i + 1] = ircounts[i] + rcounts[i]


print ('its:', its)
print ('mds:', mds)
print ('rcounts', rcounts)
print ('ircounts', ircounts)
print (res)

# the same for left ends is obvious (ilcounts[i] = i + 1)
for i in range(n):
	res[i] += ts[i] * (i + 1 - ircounts[i + 1])


print (' '.join(map(str, res)))
