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 PHP (7.4.4) (judge verdict: Wrong Answer)

<?php
$n=trim(fgets(STDIN));
$x=trim(fgets(STDIN));
$q=gmp_popcount($x);
$p01=$q+1;
$p10=$q-1;
$r01=0;
$r10=0;
$q01[0]=1;
$q10[0]=1;
$i=0;
while($i<$n){
  $r01=($r01*2)%$p01+$x[$i];
  $q01[$i+1]=$q01[$i]*2%$p01;
  if($p10>0){
    $r10=($r10*2)%$p10+$x[$i];
    $q10[$i+1]=$q10[$i]*2%$p10;
  }
  ++$i;
}

$i=0;
while($i<$n){
  $ans=1;
  if($x[$i]==0){
    $r=($r01+$q01[$n-1-$i])%$p01;
    $p=gmp_popcount($r);
  }else{
    if($p10==0){
      $p=0;
      $ans=0;
    }else{
      $r=($r10-$q10[$n-1-$i]+$p10)%$p10;
      $p=gmp_popcount($r);
    }
  }
  while($p>0){
    $r%=$p;
    $p=gmp_popcount($r);
    ++$ans;
  }
  echo "$ans\n";
  ++$i;
}