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 defaultdict, Counter
#import random
#from bisect import bisect, bisect_left
#import sys

import io, os, sys
input = io.BytesIO(os.read(0,os.fstat(0).st_size)).readline
#sys.setrecursionlimit()

def read():
    #t = int(input() )
    #for _ in range(t):
        n = int( input() )
        #n, m = map(int, input().split() )
        s = input().decode() #list(map(int, input().split() ) )
        #d = list(input().split() )
        solve(n, s)

    #r, b, d = map(int, input().split() )
    #a = list(map(int, input().split() ) )
    #q = [tuple(map(int, input().split())) for _ in range(Q)]
    #c = [tuple(map(int, input().split() ) ) for _ in range(n)]
    
def solve(n, s):
    a = []
    b = []
    for i in range(n):
        if s[i*2] == '1':
            a.append(i)
        else:
            b.append(i)
    # print(a)
    # print(b)
    # print()
    if len(a) == 0:
        sys.stdout.write("0\n")
        return
    l_a = len(a)
    l_b = len(b)
    dp = [ [n*n]*l_a for _ in range(l_b)]

    dp[0][0] = abs(a[0] - b[0])

    # for line in dp:
    #     sys.stdout.write(" ".join(map(str, line)))
    #     sys.stdout.write("\n")
    # print()

    for j in range(1, l_b):
        res = abs(b[j]-a[0])
        dp[j][0] = min(dp[j-1][0], res)
    
    for i in range(1, l_a):
        dp[i][i] = abs(b[i] - a[i]) + dp[i-1][i-1]
        for j in range(i+1, l_b):
            res = abs(b[j] - a[i]) + dp[j-1][i-1]
            dp[j][i] = min(dp[j-1][i], res)
    
    # for line in dp:
    #     sys.stdout.write(" ".join(map(str, line)))
    #     sys.stdout.write("\n")
    # print()
    sys.stdout.write(str(dp[-1][-1]))
    sys.stdout.write("\n")

if __name__ == "__main__":
    read()