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 collections import Counter, defaultdict, deque
import bisect
from sys import stdin, stdout
from itertools import repeat, permutations
import math


def inp(force_list=False):
    re = map(int, raw_input().split())
    if len(re) == 1 and not force_list:
        return re[0]
    return re

def inst():
    return raw_input().strip()

def gcd(x, y):
   while(y):
       x, y = y, x % y
   return x

MOD = 1000000007

def qmod(a, b, mod):
    res = 1
    while b:
        if b&1:
            res = (res*a)%mod
        b <<= 1
        a = (a*a)%mod
    return res

def my_main():
    kase = 1#inp()
    pans = []
    for _ in xrange(kase):
        n = inp()
        da = inp(True)
        dp = [[MOD for i in range(n+1)] for i in range(n+1)]
        dp[0] = [0] * (n+1)
        ok = 0
        for i in xrange(1, n+1):
            ok |= da[i-1]
            if not ok:
                dp[i][0] = 0
            for j in xrange(1, n+1):
                if da[i-1] == 0:
                    dp[i][j] = dp[i-1][j]
                elif da[j-1] == 1:
                    dp[i][j] = dp[i][j-1]
                else:
                    dp[i][j] = min(dp[i-1][j-1]+abs(i-j), dp[i][j-1])
        # print dp
        print dp[n][n]




    # print '\n'.join(pans)

my_main()
