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.
def solve(i, j, f, v):
    if i >= f:
        return 0
    if j >= v:
        return float('inf')

    if dp[i][j] != -1:
        return dp[i][j]
    dp[i][j] = min(abs(fill[i] - vact[j]) + solve(i + 1, j + 1, f, v), solve(i, j + 1, f, v))
    return dp[i][j]


n = int(input())
# dp = [[-1 for _ in range(n + 4)] for j in range(n + 4)]
list1 = list(map(int, input().split()))

fill = []
vact = []
f = 0
for i in range(n):
    if list1[i] == 1:
        fill.append(i)
        f = 1
    else:
        vact.append(i)
if f == 0:
    print(0)
else:
    ans = 0
    # solve(0, 0, len(fill), len(vact))
    dp = [[float('inf') for i in range(n+1)]for j in range(n+1)]


    for i in range(len(fill),n+1):
        for j in range(n+1):
            dp[i][j] = 0
    # print(dp)
    for i in range(len(fill)-1,-1,-1):
        for j in range(len(vact)-1,-1,-1):

            dp[i][j] = min(abs(vact[j]-fill[i])+dp[i+1][j+1],dp[i][j+1])

    print(dp[0][0])


