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 ll=long long;
using ull=unsigned long long;
using pii=pair<int,int>;

#define INF LONG_MAX
#define MOD 1000000007
#define rng(a) a.begin(),a.end()
#define rrng(a) a.end(),a.begin()
#define endl "\n"

int_fast64_t mpow(int_fast64_t m,int_fast64_t x,int_fast64_t n){
  using i64=int_fast64_t;
  if(n==0)return 1;
  if(n%2)return ((x%m)*(mpow(m,x,n-1)%m))%m;
  else {
    i64 half=mpow(m,x,n/2)%m;
    return (half*half)%m;
  }
}

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);

  int N;
  string X;
  cin>>N>>X;
  reverse(rng(X));

  ll bits=0;
  for(int i=0;i<N;i++)if(X[i]=='1')bits++;
  ll Bm=0,Bp=0;
  for(int i=0;i<N;i++)if(X[i]=='1'){
    if(bits-1!=0)Bm=(Bm+mpow(bits-1,2,i))%(bits-1);
    else Bm=0;
    Bp=(Bp+mpow(bits+1,2,i))%(bits+1);
  }
  reverse(rng(X));
  vector<ll>dp(200001,0);
  for(int i=1;i<=200000;i++){
    dp[i]=dp[i%__builtin_popcount(i)]+1;
  }
  for(int i=0;i<N;i++){
    ll ans=0;
    if(X[i]=='1'){
      ll m=bits-1;
      if(m!=0){
        ans=Bm;
        ans=((ans-mpow(m,2,N-i-1))%m+m)%m;
      }
      else{
        cout<<0<<endl;
        continue;
      }
    }
    else{
      ll m=bits+1;
      ans=Bp;
      ans=(ans+mpow(m,2,N-i-1))%m;
    }
    cout<<dp[ans]+1<<endl;
  }

  return 0;
}