# writer 解
# O(N log^2 N)

# にぶたんしてくれるやつ
from bisect import bisect_left, bisect_right

n = int(input())
s = input()
ans = 10**10

zero = s.count("0")
one = n-zero
if zero == 0 or zero == n:
  exit(print(0))

# 注意事項
out = [0]*n
if s[0] == "1":
  out[0] = 1
  for i in range(n-1, -1, -1):
    if s[i] == "1":
      out[i] = 1
    else:
      break

# "1" である index をとりあえずいっぱい列挙
bit = []
for i in range(n):
  if s[i] == "1":
    bit += [i, i+n, i+n+n, i+n+n+n]
bit.sort()

def f(l, ll):
  # bit[ll, ll+one) を [l, l+one) に揃えるとき、どれだけグルグル回るか

  #注意事項 
  sp_l = (l+one)%n
  sp_r = n
  if sp_l == 0:
    sp_l = n
  
  ansl = 0
  if bit[ll] <= l+n:
    # 注意事項
    # ((x-1)%n+1) で 0...n-1 ではなく 1...n にする
    if sp_l <= ((bit[ll]-1)%n+1) <= sp_r and out[bit[ll]%n] == 0:
      ansl = l+n-bit[ll]-1
    else:
      ansl = l+n-bit[ll]
  
  ansr = 0
  if l+n+one-1 < bit[ll+one-1]:
    # "右側" である要素を数えるにぶたん
    ng = -1
    ok = one
    while ng+1 != ok:
      mid = (ng+ok)//2
      if l+n+mid < bit[ll+mid]:
        ok = mid
      else:
        ng = mid
    
    # 注意事項
    flag1 = True
    flag2 = False
    x = l+n+ok
    if x%n != 0:
      x += n-(x%n)
    if x <= l+n+one-1:
      if l+n+ok <= bit[ll+ok] <= x:
        flag2 = True
      x += n
    if l+n+ok <= bit[ll+ok] <= x:
      flag1 = False
    ansr = one-ok
    ansr += flag1
    ansr -= flag2
  return max(ansl, ansr)

# 目標の区間を全探索
for l in range(n):
  # グルグル回る以前の操作回数
  anss = (-l-one)%n
  
  # 注意事項
  sp_l = (l+one)%n
  sp_r = n
  if sp_l == 0:
    sp_l = n
  
  # 何回グルグル回れば揃えられるかにぶたん
  ng = -1
  ok = n
  while ng+1 != ok:
    mid = (ng+ok)//2
    
    # 回る回数を決め打ちし、bit[ll, ll+one) の ll をできるだけ小さくする
    
    # 注意事項
    x = l+n-mid-1
    if not(sp_l <= ((x-1)%n+1) <= sp_r and out[x%n] == 0):
      x += 1
    ll = bisect_left(bit,x)
    
    if f(l, ll) <= mid:
      ok = mid
    else:
      ng = mid
  
  ans = min(ans, anss+ok*n)
print(ans)