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 Kotlin (1.3.71) (judge verdict: Accepted)

import java.io.BufferedReader
import java.io.InputStream
import java.io.InputStreamReader
import java.io.PrintWriter
import java.util.*

fun f(x: CharArray, popcount: Int): Int {
    var ret = 0L
    var pow2 = 1L
    val n = x.count()
    for (i in n - 1 downTo 0) {
        if (x[i] == '1') {
            ret += pow2
            ret %= popcount
        }
        pow2 *= 2
        pow2 %= popcount
    }
    return ret.toInt()
}

fun g(n: Long): Long {
    var popcount = 0L
    var m = n
    while (m > 0) {
        popcount += m % 2
        m /= 2
    }
    return n % popcount
}

fun PrintWriter.solve(sc: FastScanner) {
    val n = sc.nextInt()
    val a = sc.nextLine().toCharArray()
    val popcount = a.count { it == '1' }
    val minus = Array(n) { 0L }
    val plus = Array(n) { 0L }
    var pow2_m = 1L
    var pow2_p = 1L
    for (i in n - 1 downTo 0) {
        minus[i] = pow2_m
        plus[i] = pow2_p
        pow2_m *= 2
        pow2_m %= if (popcount == 1) 1 else popcount - 1
        pow2_p *= 2
        pow2_p %= popcount + 1
    }
    var sum_m = 0L
    var sum_p = 0L
    for (i in 0 until n) {
        if (a[i] == '1') {
            sum_m += minus[i]
            sum_p += plus[i]
            sum_m %= if (popcount == 1) 1 else popcount - 1
            sum_p %= popcount + 1
        }
    }
    for (i in 0 until n) {
        var count = 1
        var v = 0L
        if (a[i] == '0') {
            v = (sum_p + plus[i]) % (popcount + 1)
        }
        else {
            if (popcount == 1) {
                println(0)
                continue
            }
            v = (sum_m - minus[i] + popcount - 1) % (popcount - 1)
        }
        while (v > 0) {
            v = g(v)
            count++
        }
        println(count)
    }
}

fun main() {
    val writer = PrintWriter(System.out, false)
    writer.solve(FastScanner(System.`in`))
    writer.flush()
}

class FastScanner(s: InputStream) {
    private var st = StringTokenizer("")
    private val br = BufferedReader(InputStreamReader(s))

    fun next(): String {
        while (!st.hasMoreTokens()) st = StringTokenizer(br.readLine())

        return st.nextToken()
    }

    fun nextInt() = next().toInt()
    fun nextLong() = next().toLong()
    fun nextLine() = br.readLine()
    fun nextDouble() = next().toDouble()
    fun ready() = br.ready()
}