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 sys
def ni():
    return int(sys.stdin.readline())
def nis():
    return map(int, sys.stdin.readline().split())
def nia():
    return [*map(int, sys.stdin.readline().split())]
def ns():
    return sys.stdin.readline().rstrip()
def make_carr(dimension, value=0):
    from ctypes import c_int64, memset, sizeof, pointer
    t = c_int64
    for i in dimension[::-1]: t = t*i
    ret = t()
    if value != 0: memset(pointer(ret), value, sizeof(ret))
    return ret

from collections import deque
from functools import reduce


T = 1
for TT in range(T):
    n = ni()
    a = nia()
    seat = [i for i in range(n) if a[i]==1] # already sorted
    # dp = make_carr([len(seat)+1, n+1]) # dp[idx][here] 
    dp = [[0]*(n+1) for i in range(len(seat)+1)]
    for i in range(len(seat)+1)[::-1]:
        for j in range(n+1)[::-1]:
            if i == len(seat):
                dp[i][j] = 0
                continue
            if j == n:
                dp[i][j] = 10**15 # Here, i != len(seat)
                continue
            dp[i][j] = dp[i][j+1]
            if a[j]==0:
                cost = abs(j-seat[i])
                dp[i][j] = min(dp[i][j], dp[i+1][j+1]+cost)
    print(dp[0][0])#