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: Runtime Error)

package main

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

var m = make(map[string]string)

func main() {
	_ = readInt()
	ss := readString()

	basebitcnt := 0
	for i := 0; i < len(ss); i++ {
		if ss[i] == '1' {
			basebitcnt++
		}
	}

	for l := 0; l < len(ss); l++ {
		bitcnt := basebitcnt
		b := []byte(ss)
		if ss[l] == '0' {
			b[l] = '1'
			bitcnt++
		} else {
			b[l] = '0'
			bitcnt--
		}
		s := string(b)
		// fmt.Println("start: ", s)

		ans := 0
		for {
			if _, ok := m[s]; ok {
				for s != "0" {
					s = m[s]
					ans++
				}
				fmt.Println(ans)
				break
			}

			bitcnt := 0
			for i := 0; i < len(s); i++ {
				if s[i] == '1' {
					bitcnt++
				}
			}

			r := 0
			pre := 1
			for i := len(s) - 1; i >= 0; i-- {
				if s[i] == '1' {
					r = (r + pre) % bitcnt
				}
				pre = pre * 2 % bitcnt
			}
			next := fmt.Sprintf("%b", r)
			m[s] = next
			s = next
			ans++
			// fmt.Println(r, s)

			if s == "0" {
				fmt.Println(ans)
				break
			}
		}
	}

}

var (
	readString func() string
	readBytes  func() []byte
)

func init() {
	log.SetFlags(log.Lshortfile)
	readString, readBytes = newReadString(os.Stdin)
}

func newReadString(ior io.Reader) (func() string, func() []byte) {
	r := bufio.NewScanner(ior)
	r.Buffer(make([]byte, 1024), int(1e+11))
	r.Split(bufio.ScanWords)

	f1 := func() string {
		if !r.Scan() {
			panic("Scan failed")
		}
		return r.Text()
	}
	f2 := func() []byte {
		if !r.Scan() {
			panic("Scan failed")
		}
		return r.Bytes()
	}
	return f1, f2
}

func readInt64() int64 {
	i, err := strconv.ParseInt(readString(), 10, 64)
	if err != nil {
		panic(err.Error())
	}
	return i
}

func readInt() int {
	return int(readInt64())
}

func readFloat64() float64 {
	f, err := strconv.ParseFloat(readString(), 64)
	if err != nil {
		panic(err.Error())
	}
	return f
}

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
}

func abs(a int) int {
	if a < 0 {
		return -a
	}
	return a
}

func sum(a []int) int {
	var ret int
	for i := range a {
		ret += a[i]
	}
	return ret
}

func sumFloat64(a []float64) float64 {
	var ret float64
	for i := range a {
		ret += a[i]
	}
	return ret
}

func gcd(m, n int) int {
	for m%n != 0 {
		m, n = n, m%n
	}
	return n
}

func lcm(m, n int) int {
	return m / gcd(m, n) * n
}