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: Time Limit Exceeded)

package main

import (
	"bufio"
	"fmt"
	"math/bits"
	"os"
	"strconv"
	"strings"
)

func pow2mod(n, mod int) int {
	if n == 0 {
		return 1
	}

	ans := 1
	for i := 0; i < n; i++ {
		ans = ans * 2 % mod
	}
	return ans
}

func main() {
	N := getStdinInt()
	X := getStdin()

	cnt := 0
	for i := 0; i < len(X); i++ {
		if X[i] == '1' {
			cnt++
		}
	}
	if len(X) < N {
		for i := 0; i < N-len(X); i++ {
			X = "0" + X
		}
	}

	for n := 0; n < N; n++ {
		cnt2 := cnt
		num := 0

		if X[n] == '0' {
			cnt2++ // ビット逆転
		} else {
			cnt2-- // ビット逆転
		}

		if cnt2 <= 0 {
			fmt.Printf("%d\n", 0)
			continue
		}

		// 初回数値( mod cnt2 )
		for i := 0; i < N; i++ {
			if n == i {
				if X[i] == '0' {
					num = (num + pow2mod(N-1-i, cnt2)) % cnt2
				}
			} else {
				if X[i] == '1' {
					num = (num + pow2mod(N-1-i, cnt2)) % cnt2
				}
			}
		}

		if num <= 0 {
			fmt.Printf("%d\n", 1)
			continue
		}

		// ２回目以降
		ans := 2
		for {
			num = num % popcount(uint64(num))
			if num == 0 {
				break
			}
			ans++
		}
		fmt.Printf("%d\n", ans)
	}

}

func popcount(n uint64) int {
	return bits.OnesCount64(n)
}

var sc = bufio.NewReaderSize(os.Stdin, 1024*1024*10)

func getStdin() string {
	return readLine()
}
func getStdinInt() int {
	str := getStdin()
	rtn, _ := strconv.Atoi(str)
	return rtn
}
func getStdinInt2() (int, int) {
	list := getStdinIntArr()
	return list[0], list[1]
}
func getStdinInt3() (int, int, int) {
	list := getStdinIntArr()
	return list[0], list[1], list[2]
}
func getStdinInt64() int64 {
	str := getStdin()
	rtn, _ := strconv.ParseInt(str, 10, 64)
	return rtn
}
func getStdinInt64_2() (int64, int64) {
	list := getStdinIntArr64()
	return list[0], list[1]
}
func getStdinInt64_3() (int64, int64, int64) {
	list := getStdinIntArr64()
	return list[0], list[1], list[2]
}
func getStdinIntArr() []int {
	str := getStdin()
	list := strings.Split(str, " ")
	rtn := make([]int, len(list))
	for idx, val := range list {
		rtn[idx], _ = strconv.Atoi(val)
	}
	return rtn
}
func getStdinIntArr64() []int64 {
	str := getStdin()
	list := strings.Split(str, " ")
	rtn := make([]int64, len(list))
	for idx, val := range list {
		rtn[idx], _ = strconv.ParseInt(val, 10, 64)
	}
	return rtn
}
func readLine() string {
	buf := make([]byte, 0, 0)
	for {
		l, p, _ := sc.ReadLine()
		buf = append(buf, l...)
		if !p {
			break
		}
	}
	return string(buf)
}