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: Accepted)

#include<bits/stdc++.h>
using namespace std;
using UL = unsigned int;
using ULL = unsigned long long;
using LL = long long;
#define rep(i,n) for(UL i=0; i<(n); i++)

UL N;
UL P;
char X[200001];
UL M[200001];
UL D[200001];
UL ans[200000]={};
UL B;

UL popcount(UL x){
 x=(x&0x55555555)+((x>>1)&0x55555555);
 x=(x&0x33333333)+((x>>2)&0x33333333);
 x=(x&0x0f0f0f0f)+((x>>4)&0x0f0f0f0f);
 x=(x&0x00ff00ff)+((x>>8)&0x00ff00ff);
 x=(x&0x0000ffff)+((x>>16)&0x0000ffff);
 return x;
}

void Calc(UL m){
 M[0]=1;
 for(UL i=1; i<=N; i++) M[i]=M[i-1]*2%m;
 D[0]=0;
 for(UL i=1; i<=N; i++){
  D[i]=D[i%popcount(i)]+1;
 }
 B=0;
 for(UL i=N-1; i!=~0u; i--){
  B=B*2+(X[i]-'0');
  B%=m;
 }
}

int main() {
 scanf("%u",&N);
 scanf("%s",X);
 reverse(X,X+N);
 P=0;
 rep(i,N) if(X[i]=='1') P++;
 
 Calc(P+1);
 rep(i,N){
  if(X[i]=='0'){
   UL b=(B+M[i])%(P+1);
   ans[N-1-i]=D[b]+1;
  }
 }
 if(P!=1){
  Calc(P-1);
  rep(i,N){
   if(X[i]=='1'){
    UL b=(B+(P-1)-M[i])%(P-1);
    ans[N-1-i]=D[b]+1;
   }
  }
 }

 rep(i,N) printf("%u\n",ans[i]);

 return 0;
}