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: Runtime Error)

import java.lang.StringBuilder
import javax.print.attribute.*
import kotlin.math.*

typealias Num = Int

fun str() = readLine()!!
fun int() = str().toInt()
fun long() = str().toLong()

fun read(d:String = " ") = str().split(d)
fun <T> list(d:String = " ",f:String.()->T) = read(d).map {f(it)}
fun list() = list {toInt()}

fun <T> MutableList<T>.removeLast() = this.removeAt(this.lastIndex)

const val mod = 998244353L

fun main() {
	abc000X()
}

fun abc000X() {
	val n = int()
	val t = str()
	val bits = t.count {it=='1'}

	val high = bits+1
	val low = bits-1
	val highd = IntArray(n) {0}
	val lowd = IntArray(n) {0}
	highd[0] = 1
	lowd[0] = 1

	var highv = (t.last()-'0').toLong()
	var lowv = (t.last()-'0').toLong()

	for(i in 1 until t.length) {
		highd[i] = highd[i-1]*2%high
		lowd[i] = lowd[i-1]*2%low

		if(t[n-i-1]=='1') {
			highv = (highv+highd[i])%high
			lowv = (lowv+lowd[i])%low
		}
	}

	for(i in t.indices.reversed()) {
		val v =
				if(t[n-i-1]=='1')
					(lowv-lowd[i]+low)%low
				else
					(highv+highd[i])%high
		println(f(v)+1)
	}
}

fun f(i:Long):Int = when(i) {
	0L->0
	else->1+f(i%java.lang.Long.bitCount(i))
}