You are organizing a boxing tournament, where $n$ boxers will participate ($n$ is a power of $2$), and your friend is one of them. All boxers have different strength from $1$ to $n$, and boxer $i$ wins in the match against boxer $j$ if and only if $i$ is stronger than $j$.

The tournament will be organized as follows: $n$ boxers will be divided into pairs; the loser in each pair leaves the tournament, and $\frac{n}{2}$ winners advance to the next stage, where they are divided into pairs again, and the winners in all pairs advance to the next stage, and so on, until only one boxer remains (who is declared the winner).

Your friend really wants to win the tournament, but he may be not the strongest boxer. To help your friend win the tournament, you may bribe his opponents: if your friend is fighting with a boxer you have bribed, your friend wins even if his strength is lower.

Furthermore, during each stage you distribute the boxers into pairs as you wish.

The boxer with strength $i$ can be bribed if you pay him $a_i$ dollars. What is the minimum number of dollars you have to spend to make your friend win the tournament, provided that you arrange the boxers into pairs during each stage as you wish?


-----Input-----

The first line contains one integer $n$ ($2 \le n \le 2^{18}$) — the number of boxers. $n$ is a power of $2$.

The second line contains $n$ integers $a_1$, $a_2$, ..., $a_n$, where $a_i$ is the number of dollars you have to pay if you want to bribe the boxer with strength $i$. Exactly one of $a_i$ is equal to $-1$ — it means that the boxer with strength $i$ is your friend. All other values are in the range $[1, 10^9]$.


-----Output-----

Print one integer — the minimum number of dollars you have to pay so your friend wins.


-----Examples-----
Input
4
3 9 1 -1

Output
0
Input
8
11 -1 13 19 24 7 17 5

Output
12


-----Note-----

In the first test case no matter how you will distribute boxers into pairs, your friend is the strongest boxer and anyway wins the tournament.

In the second test case you can distribute boxers as follows (your friend is number $2$):

$1 : 2, 8 : 5, 7 : 3, 6 : 4$ (boxers $2, 8, 7$ and $6$ advance to the next stage);

$2 : 6, 8 : 7$ (boxers $2$ and $8$ advance to the next stage, you have to bribe the boxer with strength $6$);

$2 : 8$ (you have to bribe the boxer with strength $8$);

Difficulty: VERY_HARD
Tags: Dynamic programming

Complete search

Greedy algorithms
Skill types: Dynamic programming

Complete search

Greedy algorithms

Starter code:

Solutions:
import sys
from heapq import heappop, heappush
from math import log2
reader = (list(map(int, s.split())) for s in sys.stdin)

def bribe(n, a):
	if a[-1] < 0:
		return 0
	fr = a.index(-1)
	a[fr] = float('inf')
	h = int(log2(n))
	k = int(log2(fr + 1))
	total = a[-1]
	hq = []
	for r in range(h - 1, k, -1):
		[heappush(hq, a[i]) for i in range(2 ** r - 1, 2 ** (r + 1) - 1)]
		opponent = heappop(hq)
		total += opponent
	return total
(n,) = next(reader)
a = list(next(reader))
ans = bribe(n, a)
print(ans)


n = int(input())
a = list(map(int, input().split()))
me = a.index(-1)
if me == n - 1:
	print(0)
	return
for i in range(me + 1):
	a[i] = 0
dp = [[0] * n for _ in range(2)]
opt = [[0] * n for _ in range(2)]
dp[1][n - 1] = a[n - 1]
if n - 1 <= me + 1:
	ans = a[n - 1]
for i in range(2, 22):
	if 2 ** i > n:
		break
	s = i & 1
	t = s ^ 1
	(l, r) = (n // 2 ** (i - 1) - 1, n - i + 1)
	(pl, pr) = (n // 2 ** (i - 2) - 1, n - i + 2)
	opt[s][pr - 1] = dp[t][pr - 1]
	for j in range(pr - 2, pl - 1, -1):
		opt[s][j] = min(opt[s][j + 1], dp[t][j])
	for j in range(pl - 1, l - 1, -1):
		opt[s][j] = opt[s][j + 1]
	for j in range(r - 1, l - 1, -1):
		dp[s][j] = opt[s][j + 1] + a[j]
	if l <= me + 1:
		ans = min(dp[s][l:r])
print(ans)


from heapq import heappush, heappop
N = int(input())
A = [int(a) for a in input().split()]
for i in range(N):
	if A[i] < 0:
		k = i
		break
A = [0] + [0 if i < k else A[i] for i in range(N) if i != k]
ans = A.pop()
H = []
while N > 2:
	N //= 2
	for i in range(N):
		heappush(H, A.pop())
	ans += heappop(H)
print(ans)


def main():
	import sys
	from operator import itemgetter
	input = sys.stdin.readline
	N = int(input())
	K = N.bit_length() - 1
	A_raw = list(map(int, input().split()))
	flg = 1
	A = []
	for (i, a) in enumerate(A_raw):
		if flg:
			if a == -1:
				flg = 0
			else:
				A.append(0)
		else:
			A.append(A_raw[i])
	inf = 1 << 40
	dp = [[inf] * (K + 1) for _ in range(N)]
	dp[0][0] = 0
	for i in range(N):
		for j in range(K + 1):
			if dp[i][j] < inf:
				if j < K:
					dp[i + 1][j + 1] = min(dp[i + 1][j + 1], dp[i][j] + A[N - 2 - i])
				if N - 2 ** (K - j) > i:
					dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])
	print(dp[-1][-1])

def __starting_point():
	main()
__starting_point()


import sys
input = sys.stdin.readline
import heapq as hq
n = int(input())
a = list(map(int, input().split()))
frn = a.index(-1) + 1
t = n.bit_length()
if frn == n:
	print(0)
	exit()
ans = a[n - 1]
x = (frn - 1).bit_length()
q = []
hq.heapify(q)
for i in range(1, t)[::-1]:
	for j in range(2 ** (i - 1) - 1, 2 ** i - 1):
		hq.heappush(q, a[j])
	y = hq.heappop(q)
	if y >= 0:
		ans += y
	else:
		break
print(ans)


import sys
input = sys.stdin.readline
n = int(input())
A = list(map(int, input().split()))
x = A.index(-1)
for i in range(x + 1):
	A[i] = 0
B = list(reversed(A))
import heapq
H = []
ANS = 0
LIST = [0] + [n - (1 << i) + 1 for i in range(n.bit_length() - 1, 0, -1)]
for i in range(1, n.bit_length()):
	for j in range(LIST[i - 1], LIST[i]):
		heapq.heappush(H, B[j])
	x = heapq.heappop(H)
	ANS += x
print(ANS)


import heapq
pop = heapq.heappop
push = heapq.heappush
n = int(input())
a = list(map(int, input().split()))
ans = 0
pos = -1
for (i, x) in enumerate(a):
	if x == -1:
		pos = i
		break
if pos == n - 1:
	print(0)
else:
	ans += a[n - 1]
	cur = n - 2
	used = [0] * n
	used[pos] = 1
	used[n - 1] = 1
	Q = []
	i = n // 2
	while i >= 2:
		for _ in range(i):
			while used[cur] == 1:
				cur -= 1
			val = 0 if cur < pos else a[cur]
			push(Q, (val, cur))
			cur -= 1
		while True:
			(x, p) = pop(Q)
			if used[p] == 0:
				break
		used[p] = 1
		if p > pos:
			ans += a[p]
		i //= 2
	print(ans)


from sys import stdin
input = stdin.readline
n = int(input())
a = list(map(int, input().split()))
a.reverse()
inf = float('inf')
N = 19
dp = [[inf] * (N + 2) for _ in range(1 << N)]
sum = [0] * (1 << 21)
now = n
for i in range(1, 20):
	sum[i] = sum[i - 1] + now // 2
	now //= 2
for i in range(n)[::-1]:
	for j in range(N):
		if a[i] == -1:
			dp[i][j] = 0
			continue
		if i > sum[j]:
			continue
		dp[i][j] = dp[i + 1][j + 1] + a[i]
		if i < sum[j]:
			dp[i][j] = min(dp[i][j], dp[i + 1][j])
print(dp[0][0])


import sys
from array import array
import typing as Tp

def input():
	return sys.stdin.buffer.readline().decode('utf-8')

def main():
	from itertools import accumulate
	n = int(input())
	log = len(bin(n)) - 3
	cnt = list(accumulate([0] + [n >> i for i in range(1, log + 1)])) + [0]
	a = list(map(float, input().split()))
	inf = float('inf')
	dp = [0] + [inf] * (log + 1)
	for (i, v) in zip(range(1, n + 1), reversed(a)):
		if v == -1:
			print(int(min((x for x in dp if x < inf)) + 1e-08))
			exit()
		for j in range(log, -1, -1):
			if i <= cnt[j + 1]:
				dp[j + 1] = min(dp[j + 1], dp[j] + v)
			if i > cnt[j]:
				dp[j] = inf
main()


import heapq as hq
n = int(input())
a = list(map(int, input().split()))
N = 1 << 18 + 5
res = 0
heap = []
ok = [0] * N
for i in range(19):
	ok[1 << i] = 1
for i in range(n, 0, -1):
	if a[i - 1] == -1:
		break
	hq.heappush(heap, a[i - 1])
	if ok[i] == 1:
		res += hq.heappop(heap)
print(res)


from heapq import *
n = int(input())
a = list(map(int, input().split()))
done = a.index(-1)
heap = []
cost = 0
x = n // 2
position = n - 1
while position > done:
	highest = a[position]
	he = min(heap) if heap else 10 ** 10
	if highest < he:
		cost += highest
		for val in a[position - 1:position - x:-1]:
			heappush(heap, val)
		position -= x
	else:
		cost += he
		heappop(heap)
		for val in a[position:position - x:-1]:
			heappush(heap, val)
		position -= x
	x //= 2
print(cost)


from math import log2, floor

def is_power2(num):
	return num != 0 and num & num - 1 == 0

def next_two_pow(val):
	pw = 0
	while 2 ** pw <= val:
		pw = +1
	return pw
n = int(input())
arr = [int(x) for x in input().split()]
win_idx = -1
selected = []
for i in range(1, n + 1):
	val = arr[i - 1]
	if win_idx == -1:
		if val == -1:
			win_idx = i
	elif is_power2(i):
		selected.append(val)
		selected.sort()
	elif len(selected) > 0 and val < selected[-1]:
		selected.pop()
		selected.append(val)
		selected.sort()
print(sum(selected))


import math
import sys
from sys import stdin, stdout

def getMinimunCost(a, sum, i, j, dp):
	if dp[i][j] != -1:
		return dp[i][j]
	if a[j] == -1:
		return 0
	remaining = sum[i] - j
	res = -2
	if i < len(dp) - 1:
		min1 = getMinimunCost(a, sum, i + 1, j + 1, dp)
		if min1 != -2:
			res = min1 + a[j]
	if remaining > 0:
		min2 = getMinimunCost(a, sum, i, j + 1, dp)
		if res == -2:
			res = min2
		elif min2 != -2:
			res = min(res, min2)
	dp[i][j] = res
	return dp[i][j]

def getMinimunCost2(a, sum, dp):
	if a[0] == -1:
		for j in range(0, len(dp[0])):
			dp[0][j] = 0
	else:
		for j in range(0, len(dp[0])):
			dp[0][j] = a[0]
	for i in range(1, len(dp)):
		for j in range(1, len(dp[i])):
			if j <= sum[i]:
				if j > 0:
					dp[i][j] = min(dp[i][j], dp[i][j - 1])
					if i > 0:
						dp[i][j] = min(dp[i][j], dp[i - 1][j - 1] + a[j])
			else:
				dp[i][j] = min(dp[i][j], dp[i][j - 1])
	return dp[-1][-1]
try:
	n = int(stdin.readline())
	a = [int(i) for i in stdin.readline().split()]
	bcnt = int(math.log(n, 2))
	sum = []
	sum.append(0)
	x = int(n / 2)
	for i in range(1, bcnt + 1):
		sum.append(sum[i - 1] + x)
		x = int(x / 2)
	dp = [[float('inf') for i in range(n)] for i in range(bcnt + 1)]
	flag = True
	for i in range(len(a)):
		if a[i] == -1:
			flag = False
			a[i] = 0
		if flag:
			a[i] = 0
	a.reverse()
	res = getMinimunCost2(a, sum, dp)
	stdout.write(str(res))
except BaseException as e:
	print(str(e))