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.
n=int(input())
l=list(map(int,input().split()))
ans=0
l1=l.copy()
l2=l.copy()
l2=l2[::-1]
l3=l2.copy()
for i in range(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))
        elif k2==-1:
            l[k1]=2
            ans+=abs(i-k1)
        elif k1!=-1 and k2!=-1:
            if abs(k1-i)<abs(k2-i):
                l[k1]=2
                ans+=abs(k1-i)
            else:
                l[k2]=2
                ans+=abs(k2-i)
ans1=0
for i in range(n):
    if l1[i]==1:
        k1=-1
        k2=-1
        for j in range(i+1,n):
            if l1[j]==0:
                k1=j
                break
        for k in range(i-1,-1,-1):
            if l1[k]==0:
                k2=k
                break
        if k1==-1:
            l1[k2]=2
            ans1+=(abs(i-k2))
        elif k2==-1:
            l1[k1]=2
            ans1+=abs(i-k1)
        elif k1!=-1 and k2!=-1:
            if abs(k1-i)<=abs(k2-i):
                l1[k1]=2
                ans1+=abs(k1-i)
            else:
                l1[k2]=2
                ans1+=abs(k2-i)
ans2=0
for i in range(n):
    if l2[i]==1:
        k1=-1
        k2=-1
        for j in range(i+1,n):
            if l2[j]==0:
                k1=j
                break
        for k in range(i-1,-1,-1):
            if l2[k]==0:
                k2=k
                break
        if k1==-1:
            l2[k2]=2
            ans2+=(abs(i-k2))
        elif k2==-1:
            l2[k1]=2
            ans2+=abs(i-k1)
        elif k1!=-1 and k2!=-1:
            if abs(k1-i)<=abs(k2-i):
                l2[k1]=2
                ans2+=abs(k1-i)
            else:
                l2[k2]=2
                ans2+=abs(k2-i)
ans3=0
for i in range(n):
    if l3[i]==1:
        k1=-1
        k2=-1
        for j in range(i+1,n):
            if l3[j]==0:
                k1=j
                break
        for k in range(i-1,-1,-1):
            if l3[k]==0:
                k2=k
                break
        if k1==-1:
            l3[k2]=2
            ans3+=(abs(i-k2))
        elif k2==-1:
            l3[k1]=2
            ans3+=abs(i-k1)
        elif k1!=-1 and k2!=-1:
            if abs(k1-i)<abs(k2-i):
                l3[k1]=2
                ans3+=abs(k1-i)
            else:
                l3[k2]=2
                ans3+=abs(k2-i)
print(min(ans,ans1,ans2,ans3))