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 Scala (2.13.1) (judge verdict: Time Limit Exceeded)

object Main {
  def main(args: Array[String]): Unit = {
    val n_length = scala.io.StdIn.readInt
    val n        = scala.io.StdIn.readLine

    var map: Map[ String, Int ] = Map.empty

    ( 0 until n_length ).foreach{ i =>
      val change_value = if( n( i ) == '0' ) '1' else '0'

      var check_n = n.updated( i, change_value )
      val tmp = check_n
      var count = 0

      while( check_n != "0" ) {
        if( map.contains( check_n ) ) {
          count += map( check_n )
          check_n = "0"
        }
        else {
          count += 1
          check_n = calcBinaryDiv( check_n, popCount( check_n ).toBinaryString )
        }
      }
      map += ( tmp -> count )

      println( count )
    }
  }

  def calcBinaryDiv( n: String, divided_n: String ): String = {
    val check = n.split( "" )
    var x = ""
    var answer = ""

    check.foreach{ i =>
      x += i
      val calc_x = java.lang.Integer.parseInt( x, 2 ) - java.lang.Integer.parseInt( divided_n, 2 )
      if( calc_x >= 0 ) {
        x = calc_x.toBinaryString
        answer += "1"
      }
      else if( answer != "" ) {
        answer += "0"
      }
    }

    return java.lang.Integer.parseInt( x, 2 ).toBinaryString
  }

  def popCount( n: String ): Int = {
    val check = n.split( "" ).map{ _.toInt }
    var count = 0
    check.foreach{ i =>
      if( i == 1 ) count += 1
    }
    return count
  }
}