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 <bits/stdc++.h>
#define rep(i,n) for(int i = 0; i < (int)(n); i++)
#define rrep(ri,n) for(int ri = (int)(n-1); ri >= 0; ri--)
#define rep2(i,x,n) for(int i = (int)(x); i < (int)(n); i++)
#define rrep2(ri,x,n) for(int ri = (int)(n-1); ri >= (int)(x); ri--)
#define repit(itr,x) for(auto itr = x.begin(); itr != x.end(); itr++)
#define rrepit(ritr,x) for(auto ritr = x.rbegin(); ritr != x.rend(); ritr++)
#define ALL(x) x.begin(), x.end()
using ll = long long;
using namespace std;

int fc(int i){
  int co = 0;
  int j = i;
  while(j != 0){
    if(j % 2 == 1) co++;
    j /= 2;
  }
  int nx = i % co;
  if(nx == 0) return 1;
  return 1 + fc(nx);
}

int st(const string &x, int &co, int index){
  int n = x.size();
  co = 0;
  static vector<int> xtb(n, -1), xtp(n, -1), xtm(n, -1);
  if(xtp.at(index) != -1 || xtm.at(index) != -1){
    if(xtp.at(index) != -1) return xtp.at(index);
    return xtm.at(index);
  }
  vector<int> powerp(n), powerm(n);
  rep(i, n){
    if(x.at(i) == '1') co++;
  }
  int nowm = 0, nowp = 0;
  rrep(i, n){
    if(i == n-1){
      powerm.at(i) = powerp.at(i) = 1;
    }else{
      if(co > 1) powerm.at(i) = ((ll)powerm.at(i+1)*2) % (co-1);
      powerp.at(i) = ((ll)powerp.at(i+1)*2) % (co+1);
    }
  }
  rep(i, n){
    nowp *= 2;
    nowp += x.at(i) - '0';
    nowp %= co+1;
    if(co > 1){
      nowm *= 2;
      nowm += x.at(i) - '0';
      nowm %= co-1;
    }
  }
  rep(i, n){
    if(x.at(i) == '0'){
      int add = nowp;
      add += powerp.at(i);
      add %= (co+1);
      xtp.at(i) = add;
    }else{
      if(co <= 1) xtm.at(i) = -2;
      else{
        int add = nowm;
        add -= powerm.at(i);
        while(add < 0) add += (co-1);
        add %= (co-1);
        xtm.at(i) = add;
      }
    }
  }
  return -1;
}

int main(){
  int n;
  cin >> n;
  string x;
  cin >> x;
  int co;
  st(x, co, 0);
  rep(i, n){
    int r = st(x, co, i);
    if(r == 0){
      cout << 1 << endl;
      continue;
    }
    if(r == -2){
      cout << 0 << endl;
      return 0;
    }
    int ans = 1 + fc(r);
    cout << ans << endl;
  }
  return 0;
}