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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <math.h>
#include <stdlib.h>
int popcount(int n);

void main(void){
    int X,N,pc=0;
    scanf("%d",&N);
    int by[N];
    int i = 0;
    char ch[N];
    scanf("%s",ch);

    //charで読み込んだ01の二進数をintにしてby[]に格納
    for(int i = 0;i < N;i++){
        if(ch[i] == '0')by[i] = 0;
        else {
            by[i] = 1;
            pc++;//1の時カウント
        }
        
    }
    //この時点で最初に与えられたN桁の二進数の1の数がpcに格納される
    int Fpc = pc;//初期の1の数をメモ

    //pow[n]に2^nの値を格納
    int pow[N];
    pow[N-1] = 1;
    for(int k = N-2;k >= 0;k--){
        pow[k] = pow[k+1]*2;
    }

    //与えられたN桁の二進数を10進数に変換する
    int n = 0;
    for(int j = 0;j < N;j++){
            n += by[j]*pow[j];
        }
    int Fn = n;//nの初期値をメモ
    
    //ここからf(Xi)を計算する処理
    for(int i = 0;i < N;i++){
        int count = 0;
        pc = Fpc;
        n = Fn;

        //i番目が0の時反転させて1にその分1の数が増え、nも増加するためn = n + 2^(N-i)
        if(by[i] == 0){
            pc++;
            n += pow[i];
        }
        //i番目が1の時反転させて0に1が減るため、n = n - 2^(N-i)
        if(by[i] == 1){
            pc--;
            n -= pow[i];
        }

        //nをpcで割った余りを新しいnとし、新しいnの2進表記の1の数を求める。
        while (n != 0){
            if(n == 0 || pc == 0)break;
            n = n%pc;
            count++;
            pc = popcount(n);
        }

        printf("%d\n",count);
          
    }
}

int popcount(int n){
    int B;
    int sum = 0;
    while(n != 0){
        B = n%2;
        n = (n-B)/2;
        if(B == 1)sum++;
    }
    return sum;
}