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 Go (1.14.1) (judge verdict: Accepted)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc, wr = bufio.NewScanner(os.Stdin), bufio.NewWriter(os.Stdout)

func scanString() string   { sc.Scan(); return sc.Text() }
func scanRunes() []rune    { return []rune(scanString()) }
func scanInt() int         { a, _ := strconv.Atoi(scanString()); return a }
func scanInt64() int64     { a, _ := strconv.ParseInt(scanString(), 10, 64); return a }
func scanFloat64() float64 { a, _ := strconv.ParseFloat(scanString(), 64); return a }

func scanInts(n int) []int {
	res := make([]int, n)
	for i := 0; i < n; i++ {
		res[i] = scanInt()
	}
	return res
}

func debug(a ...interface{}) { fmt.Fprintln(os.Stderr, a...) }

func abs(a int) int {
	if a < 0 {
		return -a
	}
	return a
}
func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}
func max(a, b int) int {
	if a > b {
		return a
	}
	return b
}

//•*¨*•.¸¸♪main•*¨*•.¸¸♪(　-ω-)ノ　(　・ω・)
func main() {
	defer wr.Flush()
	sc.Split(bufio.ScanWords)
	sc.Buffer(make([]byte, 10000), 1001001)

	n := scanInt()
	x := scanString()

	memo := make([]int, n+1)
	for i := 0; i < n; i++ {
		c := 0
		for b := i + 1; b != 0; b >>= 1 {
			if b&1 == 1 {
				c++
			}
		}
		memo[i+1] = memo[(i+1)%c] + 1
	}

	cn := 0
	for _, s := range x {
		if s == '1' {
			cn++
		}
	}

	a1, a2 := 0, 0
	p1, p2 := 1, 1
	for i := 0; i < n; i++ {
		if x[n-1-i] == '1' {
			if cn != 1 {
				a1 = (a1 + p1) % (cn - 1)
			}
			a2 = (a2 + p2) % (cn + 1)
		}
		if cn != 1 {
			p1 = p1 * 2 % (cn - 1)
		}
		p2 = p2 * 2 % (cn + 1)
	}

	ans := make([]int, n)
	p1, p2 = 1, 1
	for i := 0; i < n; i++ {
		c := cn
		a := 0
		if x[n-1-i] == '1' {
			if cn != 1 {
				c--
				a = (a1 - p1%c + c) % c
			} else {
				ans[n-1-i] = 0
				continue
			}
		} else {
			c++
			a = (a2 + p2%c) % c
		}
		ans[n-1-i] = memo[a] + 1
		if cn != 1 {
			p1 = p1 * 2 % (cn - 1)
		}
		p2 = p2 * 2 % (cn + 1)
	}

	for i := 0; i < n; i++ {
		fmt.Fprintln(wr, ans[i])
	}
}