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<iostream>
#include<iomanip>
#include<cmath>
#include<string>
#include<cstring>
#include<vector>
#include<list>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#include<stack>
using namespace std;
typedef long long ll;
#define fi first
#define se second
#define mp make_pair
#define rep(i, n) for(int i=0;i<n;++i)
#define rrep(i, n) for(int i=n;i>=0;--i)
const int inf=1e9+7;
const ll mod=1e9+7;
const ll mod1=998244353;
const ll big=1e18;
const double PI=2*asin(1);

int ans[200005];

int main() {
  int N;
  string X;
  cin>>N>>X;
  int num;
  for(int i=1;i<200005;++i) {
    num = 0;
    for(int j=0;j<20;++j) {
      if((i&(1<<j))!=0) num++;
    }
    ans[i] = 1 + ans[i%num];
  }
  num = 0;
  for(int i=0;i<N;++i) {
    if(X[i]=='1') num++; 
  }
  if(num==0) {
    for(int i=0;i<N;++i) cout<<1<<endl;
    return 0;
  }
  if(num==1) {
    for(int i=0;i<N;++i) {
      if(X[i]=='1') {
        cout<<0<<endl;
      }
      else {
        if(i==N-1) {
          cout<<2<<endl;
        }
        else {
          cout<<1<<endl;
        }
      }
    }
    return 0;
  }
  int amari1=0, amari2=0;
  int two1=1, two2=1;
  for(int i=N-1;i>=0;--i) {
    if(X[i]=='1') {
      amari1 += two1%(num+1);
      amari2 += two2%(num-1);
      amari1 %= num+1;
      amari2 %= num-1;
    }
    two1 *= 2;
    two1 %= num+1;
    two2 *= 2;
    two2 %= num-1;
  }
  int ans1[N];
  two1 = 1;
  two2 = 1;
  for(int i=N-1;i>=0;--i) {
    if(X[i]=='1') { // amari2
      ans1[i] = ans[(amari2-two2+(num-1))%(num-1)]+1;
    }
    else { // amari1
      ans1[i] = ans[(amari1+two1)%(num+1)]+1;
    }
    two1 *= 2;
    two2 *= 2;
    two1 %= num+1;
    two2 %= num-1;
  }
  for(int i=0;i<N;++i) cout<<ans1[i]<<endl;
}