Problem:
Score : 400 points

Problem Statement

Let \mathrm{popcount}(n) be the number of 1s in the binary representation of n.
For example, \mathrm{popcount}(3) = 2, \mathrm{popcount}(7) = 3, and \mathrm{popcount}(0) = 0.

Let f(n) be the number of times the following operation will be done when we repeat it until n becomes 0: "replace n with the remainder when n is divided by \mathrm{popcount}(n)." (It can be proved that, under the constraints of this problem, n always becomes 0 after a finite number of operations.)

For example, when n=7, it becomes 0 after two operations, as follows:

\mathrm{popcount}(7)=3, so we divide 7 by 3 and replace it with the remainder, 1.

\mathrm{popcount}(1)=1, so we divide 1 by 1 and replace it with the remainder, 0.

You are given an integer X with N digits in binary.
For each integer i such that 1 \leq i \leq N, let X_i be what X becomes when the i-th bit from the top is inverted.
Find f(X_1), f(X_2), \ldots, f(X_N).

Constraints

1 \leq N \leq 2 \times 10^5

X is an integer with N digits in binary, possibly with leading zeros.

Input

Input is given from Standard Input in the following format:

N
X

Output

Print N lines. The i-th line should contain the value f(X_i).

Sample Input 1

3
011

Sample Output 1

2
1
1

X_1 = 7, which will change as follows: 7 \rightarrow 1 \rightarrow 0. Thus, f(7) = 2.

X_2 = 1, which will change as follows: 1 \rightarrow 0. Thus, f(1) = 1.

X_3 = 2, which will change as follows: 2 \rightarrow 0. Thus, f(2) = 1.

Sample Input 2

23
00110111001011011001110

Sample Output 2

2
1
2
2
1
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
2
1
3

Solution in Ruby (2.7.1) (judge verdict: Wrong Answer)

def modpow(a, n, mod)
  if n > 0 && mod > 1
    exp = Math.log(n, 2).floor
    rep = modpow_cl(a, n, exp, mod)
    return rep[0]
  elsif mod == 1
    return 0
  else
    return 1
  end
end

def modpow_cl(a, n, exp, mod)
  if exp < 0
    rep = [1, a]
  else
    if n >= 2**exp
      rep = modpow_cl(a, n-2**exp, exp-1, mod)
      rep[0] = rep[0]*rep[1] % mod
    else
      rep = modpow_cl(a, n, exp-1, mod)
    end
    rep[1] = rep[1]**2 % mod
  end
  return rep
end

def popcntRec(num, cnt)
  return cnt if num == 1
  maxexp = Math.log(num, 2).floor
  pcnt = 0
  tmp = num
  for i in 0..maxexp
    exp = maxexp - i
    ifdiv = tmp / (2**exp)
    tmp -= ifdiv * (2**exp)
    pcnt += ifdiv
  end
  if num % pcnt == 0
    return cnt
  else
    return popcntRec(num % pcnt, cnt+1)
  end
end

n = gets.to_i
x = gets.split("").map(&:to_i)

cnt1 = 0
x.each{|xe| cnt1 += 1 if xe == 1}

modsLar = Array.new(n,0)
modsSma = Array.new(n,0)

for i in 0..(n-1)
  modsLar[i] = modpow(2, n-1-i, cnt1+1)
  modsSma[i] = modpow(2, n-1-i, cnt1-1) if cnt1 > 1
end

sumorgnLar = 0
sumorgnSma = 0
for i in 0..(n-1)
  sumorgnLar = (sumorgnLar + modsLar[i]) if x[i] == 1
  sumorgnSma = (sumorgnSma + modsSma[i]) if x[i] == 1
end

ans = Array.new(n,0)
for i in 0..(n-1)
  if x[i] == 0
    modrslt = (sumorgnLar + modsLar[i]) % (cnt1+1)
    if modrslt == 0
      ans[i] = 1
    else
      ans[i] = popcntRec(modrslt, 2)
    end
  else
    if cnt1 > 1
      modrslt = (sumorgnSma - modsSma[i]) % (cnt1-1)
      if modrslt == 0
        ans[i] = 1
      else
        ans[i] = popcntRec(modrslt, 2)
      end
    else
      ans[i] = 1
    end
  end
end

for i in 0..(n-1)
  puts ans[i]
end