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 io
import os

from bisect import bisect_left, bisect_right
from collections import Counter, defaultdict, deque
from heapq import heappush, heappop, heapify
from math import gcd, inf


def solve(N, A):
    assert len(A) == N
    # Want things jumping left and things jumping right to never jump over each other
    # This means their relative ordering after moving should remain the same

    occupied = deque()
    free = deque()
    for i, x in enumerate(A):
        if x:
            occupied.append(i)
        else:
            free.append(i)
    O = len(occupied)
    F = len(free)

    cache = [-1] * ((O + 1) * (F + 1))

    def dp(i, j):
        # best way to assign occupied[:i] to free[:j]
        key = i * (F + 1) + j
        if cache[key] == -1:
            if i == 0:
                ret = 0
            elif j == 0:
                assert i > 0
                ret = inf
            else:
                assert i > 0
                assert j > 0
                ret = min(
                    abs(occupied[i - 1] - free[j - 1]) + dp(i - 1, j - 1), dp(i, j - 1)
                )
            cache[key] = ret

        return cache[key]

    for i in range(O + 1):
        for j in range(F + 1):
            dp(i, j)

    ret = dp(O, F)
    return ret


if False:
    assert solve(8, [1, 1, 0, 0, 1, 1, 0, 0]) == 8

    assert (
        solve(
            100,
            list(
                map(
                    int,
                    "0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 1 0 0 0 0 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0".split(),
                )
            ),
        )
        == 10
    )

if __name__ == "__main__":
    input = io.BytesIO(os.read(0, os.fstat(0).st_size)).readline

    TC = 1
    for tc in range(1, TC + 1):
        (N,) = [int(x) for x in input().split()]
        A = [int(x) for x in input().split()]
        ans = solve(N, A)
        print(ans)
