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(l,ans,ind):
    if ind==len(l):
        return ans
    for i in range(ind,n):
        if l[i] == 1:
            k1 = -1
            k2 = -1
            for j in range(i + 1, n):
                if l[j] == 0:
                    k1 = j
                    break
            for k in range(i - 1, -1, -1):
                if l[k] == 0:
                    k2 = k
                    break
            if k1 == -1:
                l[k2] = 2
                ans += (abs(i - k2))
                return solve(l,ans,i+1)
            elif k2 == -1:
                l[k1] = 2
                ans += abs(i - k1)
                return solve(l, ans, i + 1)
            elif k1 != -1 and k2 != -1:
                if abs(k1 - i) < abs(k2 - i):
                    l[k1] = 2
                    ans += abs(k1 - i)
                    return solve(l,ans,i+1)
                elif abs(k1-i)>abs(k2-i):
                    l[k2] = 2
                    ans += abs(k2 - i)
                    return solve(l, ans, i + 1)
                else:
                    l1=l.copy()
                    l1[k1]=2
                    l[k2]=2
                return min(solve(l,ans+abs(i-k2),i+1),solve(l1,ans+abs(i-k1),i+1))
    return ans

n=int(input())
l=list(map(int,input().split()))
print(solve(l,0,0))