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 _typeshed import SupportsKeysAndGetItem
import sys
#sys.stdout=open("output.txt","w")
####### GLOBAL ###############
MOD=1000000007
no=lambda:print("NO")
yes=lambda:print("YES")
_1=lambda:print(-1)
ari=lambda:[int(_) for _ in input().split()]
cin=lambda:int(input())
cis=lambda:input()
show=lambda x: print(x)
########### END #########
######
test_case=1
#test_case=int(input())
######
def ans():
    n=cin()
    arr=ari()
    visit=[0]*n
    for i in range(n):
        if arr[i]==1:
            visit[i]=-1
    cnt=0
    def dfs(i,curr):
        # nonlocal n
        # if curr>=n or curr<0:
        #     return 99999,n-1
        # if arr[curr]==0 and visit[curr]!=-1:
        #     return  abs(curr-i),curr
        # l,ind=dfs(i,curr+1)
        # r,ind=dfs(i,curr-1)
        # if l>r:
        #     return r,ind
        # return l,ind
        left=99999
        right=99999
        for it in range(i,n):
            if arr[it]==0 and visit[it]!=-1:
                left=it
                break
        for j in range(i,-1,-1):
            if arr[j]==0 and visit[j]!=-1:
                right=j
                break
        if left==9999:
            return  abs(curr-right),right
        if right==9999:
            return  abs(curr-left),left
        if abs(curr-right)>abs(curr-left):
            return abs(curr-left),left
        return abs(curr-right),right
            
    for i in range(n):
        if arr[i]==1:
            ok,ini=dfs(i,i)
            #print(ini,i)
           # print(ini)
            visit[ini]=-1
           # print(visit)
            cnt+=ok
    print(cnt)
    
    
    return
    
        
for _ in range(test_case):
    ans()