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 os
import sys
from bisect import bisect_right
from collections import Counter, defaultdict, deque
from heapq import *
from io import BytesIO, IOBase
from math import gcd, inf, sqrt, ceil
def lcm(a, b):
    return (a * b) // gcd(a, b)
'''
    mod = 10 ** 9 + 7
    fac = [1]
    for i in range(1, 2 * 10 ** 5 + 1):
        fac.append((fac[-1] * i) % mod)
    fac_in = [pow(fac[-1], mod - 2, mod)]
    for i in range(2 * 10 ** 5, 0, -1):
        fac_in.append((fac_in[-1] * i) % mod)
    fac_in.reverse()
    def comb(a, b):
        if a < b:
            return 0
        return (fac[a] * fac_in[b] * fac_in[a - b]) % mod
'''
#MAXN = 10000004
# spf = [0 for i in range(MAXN)]
# adj = [[] for i in range(MAXN)]
def sieve():
    global spf, adj, MAXN
    spf[1] = 1
    for i in range(2, MAXN):
        spf[i] = i

    for i in range(2, MAXN):
        if i * i > MAXN:
            break
        if (spf[i] == i):
            for j in range(i * i, MAXN, i):
                if (spf[j] == j):
                    spf[j] = i
def getdistinctFactorization(n):
    global adj, spf, MAXN
    for i in range(1, n + 1):
        index = 1
        x = i
        if (x != 1):
            adj[i].append(spf[x])
        x = x // spf[x]
        while (x != 1):
            if (adj[i][index - 1] != spf[x]):
                adj[i].append(spf[x])
                index += 1
            x = x // spf[x]
def printDivisors(n):
    i = 2
    z = [1, n]
    while i <= sqrt(n):
        if (n % i == 0):
            if (n / i == i):
                z.append(i)
            else:
                z.append(i)
                z.append(n // i)
        i = i + 1
    return z
def create(n, x, f):
    pq = len(bin(n)[2:])
    if f == 0:
        tt = min
    else:
        tt = max
    dp = [[inf] * n for _ in range(pq)]
    dp[0] = x
    for i in range(1, pq):
        for j in range(n - (1 << i) + 1):
            dp[i][j] = tt(dp[i - 1][j], dp[i - 1][j + (1 << (i - 1))])
    return dp
def enquiry(l, r, dp, f):
    if l > r:
        return inf if not f else -inf
    if f == 1:
        tt = max
    else:
        tt = min
    pq1 = len(bin(r - l + 1)[2:]) - 1
    return tt(dp[pq1][l], dp[pq1][r - (1 << pq1) + 1])
def SieveOfEratosthenes(n):
    prime = [True for i in range(n + 1)]
    p = 2
    while (p * p <= n):
        if (prime[p] == True):
            for i in range(p * p, n + 1, p):
                prime[i] = False
        p += 1
    x = []
    for i in range(2, n + 1):
        if prime[i]:
            x.append(i)
    return x

def main():
    n=int(input())
    a=[int(X) for X in input().split()]
    x=[]
    y=[]
    for i in range(n):
        if a[i]==1:
            x.append(i)
        else:
            y.append(i)
    dp=[[inf for i in range(len(x)+1)]for j in range(len(y)+1) ]
    dp[0][0]=0
    for i in range(len(y)):
        for j in range(len(x)+1):
            dp[i+1][j]=min(dp[i+1][j],dp[i][j])
            if (j<len(x)):
                dp[i+1][j+1]=min(dp[i+1][j+1],dp[i][j]+abs(y[i]-x[j]))
    print(dp[len(y)][len(x)])
# Fast IO Region
BUFSIZE = 8192
class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")

if __name__ == '__main__':
    main()
