There are n armchairs, numbered from 1 to n from left to right. Some armchairs are occupied by people (at most one person per armchair), others are not. The number of occupied armchairs is not greater than n/2.

For some reason, you would like to tell people to move from their armchairs to some other ones. If the i-th armchair is occupied by someone and the j-th armchair is not, you can tell the person sitting in the i-th armchair to move to the j-th armchair. The time it takes a person to move from the i-th armchair to the j-th one is |i - j| minutes. You may perform this operation any number of times, but these operations must be done sequentially, i. e. you cannot tell a person to move until the person you asked to move in the last operation has finished moving to their destination armchair.

You want to achieve the following situation: every seat that was initially occupied must be free. What is the minimum time you need to do it?

Input

The first line contains one integer n (2 ≤ n ≤ 5000) — the number of armchairs.

The second line contains n integers a_1, a_2, ..., a_n (0 ≤ a_i ≤ 1). a_i = 1 means that the i-th armchair is initially occupied, a_i = 0 means that it is initially free. The number of occupied armchairs is at most n/2.

Output

Print one integer — the minimum number of minutes you have to spend to achieve the following situation: every seat that was initially occupied must be free.

Examples

Input


7
1 0 0 1 0 0 1


Output


3


Input


6
1 1 1 0 0 0


Output


9


Input


5
0 0 0 0 0


Output


0

Note

In the first test, you can perform the following sequence:

  1. ask a person to move from armchair 1 to armchair 2, it takes 1 minute; 
  2. ask a person to move from armchair 7 to armchair 6, it takes 1 minute; 
  3. ask a person to move from armchair 4 to armchair 5, it takes 1 minute. 



In the second test, you can perform the following sequence:

  1. ask a person to move from armchair 1 to armchair 4, it takes 3 minutes; 
  2. ask a person to move from armchair 2 to armchair 6, it takes 4 minutes; 
  3. ask a person to move from armchair 4 to armchair 5, it takes 1 minute; 
  4. ask a person to move from armchair 3 to armchair 4, it takes 1 minute. 



In the third test, no seat is occupied so your goal is achieved instantly.
import sys
import math
import heapq
import bisect
from collections import Counter
from collections import defaultdict
from io import BytesIO, IOBase
import string


class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        import os
        self.os = os
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None
        self.BUFSIZE = 8192

    def read(self):
        while True:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = self.os.read(self._fd, max(self.os.fstat(self._fd).st_size, self.BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            self.os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")


def get_int():
    return int(input())


def get_ints():
    return list(map(int, input().split(' ')))


def get_int_grid(n):
    return [get_ints() for _ in range(n)]


def get_str():
    return input().strip()


def get_strs():
    return get_str().split(' ')


def flat_list(arr):
    return [item for subarr in arr for item in subarr]


def yes_no(b):
    if b:
        return "YES"
    else:
        return "NO"


def binary_search(good, left, right, delta=1, right_true=False):
    """
    Performs binary search
    ----------
    Parameters
    ----------
    :param good: Function used to perform the binary search
    :param left: Starting value of left limit
    :param right: Starting value of the right limit
    :param delta: Margin of error, defaults value of 1 for integer binary search
    :param right_true: Boolean, for whether the right limit is the true invariant
    :return: Returns the most extremal value interval [left, right] which is good function evaluates to True,
            alternatively returns False if no such value found
    """

    limits = [left, right]
    while limits[1] - limits[0] > delta:
        if delta == 1:
            mid = sum(limits) // 2
        else:
            mid = sum(limits) / 2
        if good(mid):
            limits[int(right_true)] = mid
        else:
            limits[int(~right_true)] = mid
    if good(limits[int(right_true)]):
        return limits[int(right_true)]
    else:
        return False


def prefix_sums(a):
    p = [0]
    for x in a:
        p.append(p[-1] + x)
    return p


def solve_a():
    k = get_int()
    g = math.gcd(k, 100 - k)
    return k // g + (100 - k) // g


def solve_b():
    n = get_int()
    a = get_ints()
    if a == list(range(1, n + 1)):
        return 0
    elif a[0] == 1 or a[-1] == n:
        return 1
    elif a[0] == n and a[-1] == 1:
        return 3
    else:
        return 2


def solve_c():
    n, m = get_ints()
    x = get_ints()
    d = get_strs()


def solve_d():
    n = get_int()
    a = get_ints()
    p = [i for (i, x) in enumerate(a) if x == 1]
    g = [i for (i, x) in enumerate(a) if x == 0]

    m = len(p)
    k = len(g)

    j = 0
    best = 0
    for i in range(m):
        lb = float('inf')
        best_j = -1
        while k - j >= m - i:
            if abs(p[i] - g[j]) < lb:
                lb = abs(p[i] - g[j])
                best_j = j
            j += 1
        best += lb
        j = best_j + 1

    return best





print(solve_d())

