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.
from sys import stdin, stdout
from collections import *
from math import gcd, ceil, floor


def st():
    return list(stdin.readline().strip())


def li():
    return list(map(int, stdin.readline().split()))


def mp():
    return map(int, stdin.readline().split())


def inp():
    return int(stdin.readline())


def pr(n):
    return stdout.write(str(n) + "\n")


mod = 1000000007
INF = float('inf')
Y = "YES"
N = "NO"


def solve():
    # solve here
    N = int(input())
    c = li()
    pos = []
    per = []
    for i in range(N):
        if c[i] == 0:
            pos.append(i)
        else:
            per.append(i)

    n = len(pos)
    m = len(per)
    dp = [[INF for i in range(m + 1)] for j in range(n + 1)]
    dp[0][0] = 0
    for i in range(n):
        for j in range(m):
            dp[i + 1][j] = min(dp[i + 1][j], dp[i][j])
            dp[i + 1][j + 1] = min(dp[i + 1][j + 1],
                                   dp[i][j] + abs(pos[i] - per[j]))
    print(min(dp[i][m] for i in range(n + 1)))


for test in range(1):
    solve()
