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: Runtime Error)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#define ll long long

ll n;
char x[212345];
int f[212345];

int minus[212345];
int plus[212345];


int popc(int d){
  int sum=0;
  int tmp=d;
  while(tmp){
    if(tmp & 1) sum++;
    tmp >>= 1;}
  return sum;
  }


void compf(void){
  for(int i=1;i<200000;i++){
    int p = popc(i);
    f[i]=f[i%p]+1;}
}

int main(){
  //printf("dd\n");
  scanf("%lld",&n);
  scanf("%s",x);
  compf();

  ll p=0;
  for(int i=0;i<n;i++){
    if(x[i]=='1') p++;}
  

  ll sum0=0,summ=0,sump=0;

  ll tmp=1;
  /*
  for(int j=n-1;j>=0;j--){
    tmp%=p;
    a0[j]=tmp;
    tmp = tmp*2;}*/
  tmp=1;
  for(int j=n-1;j>=0;j--){
    tmp%=p-1;
    minus[j]=tmp;
    tmp = tmp*2;}
  tmp=1;
  for(int j=n-1;j>=0;j--){
    tmp%=p+1;
    plus[j]=tmp;
    tmp = tmp*2;}
  for(int j=n-1;j>=0;j--){
    if(x[j]=='1'){
      //sum0 = (sum0+a0[j])%p;
      summ = (summ+minus[j])%(p-1);
      sump = (sump+plus[j])%(p+1);}
    }
    

  for(int i=0;i<n;i++){
    if(x[i]=='0'){
      ll tmp2 = (sump+plus[i])%(p+1);
      printf("%d\n",f[tmp2]+1);}
    else{
      ll tmp3 = (summ+(p-1)-minus[i])%(p-1);
      printf("%d\n",f[tmp3]+1);}
  }

  //printf("%d\n",popc(3));
  
  
  return 0;
}