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
input = stdin.readline
from heapq import heapify,heappush,heappop,heappushpop
from collections import defaultdict as dd, deque as dq,Counter as C
from math import factorial as f ,ceil,gcd,sqrt,log
from bisect import bisect_left as bl ,bisect_right as br
from itertools import combinations as c,permutations as p
from math import factorial as f ,ceil,gcd,sqrt,log,inf
mi = lambda : map(int,input().split())
ii = lambda: int(input())
li = lambda : list(map(int,input().split()))
mati = lambda  r : [ li() for _ in range(r)]
lcm = lambda a,b : (a*b)//gcd(a,b) 




def solve():
    n=ii()
    arr=li()
    arr1=[0]
    arr2=[0]
    for x in range(n):
        if arr[x]==1:
            arr1.append(x+1)
        else:
            arr2.append(x+1)
    #print(arr2)
    dp=[[0 for y in range(len(arr1)+5)] for x in range(len(arr2)+5)]
    s1,s2=len(arr1),len(arr2)
    for x in range(s1):
        for y in range(s2):
            dp[x][y]=inf
    for x in range(s2):
        dp[0][x]=0
    ans=inf
    #print(dp)
    for x in range(1,s1):
        for y in range(1,s2):
           # print(x,y)
            dp[x][y]=min(dp[x][y-1],dp[x-1][y-1]+abs(arr1[x]-arr2[y]))
            if x==s1-1:
                ans=min(ans,dp[x][y])
    print(ans)  



for _ in range(1):
    solve()