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>
using namespace std;
#ifdef _DEBUG
  #include "_DEBUG.hpp"
#endif
#define int long long
const int INF = 1LL << 60;

int mod_pow(int n, int p, int MOD){
  int r = 1;
  for(; p > 0; p >>= 1){
    if(p & 1LL) r = (r * n) % MOD;
    n = (n * n) % MOD;
  }
  return r; // r = n^p % MOD
}

int f(int x, int b){
  if(x == 0) return 0LL;
  int cnt = 0;
  int base = 1;
  while(x != 0){
    if(x % (base * b) == 0){

    }else{
      x -= base;
      cnt++;
    }
    base *= b;
  }
  return cnt;
}

signed main(){

  int n; cin >> n;
  string s; cin >> s;

  reverse(s.begin(), s.end());
  int bit = 0;
  for(int i = 0; i < n; i++){
    bit += s[i] == '1';
  }

  int r1 = 0, r2 = 0;
  for(int i = 0; i < n; i++){
    if(s[i] == '1') r1 = (r1 + mod_pow(2, i, (bit+1))) % (bit+1); 
    if(s[i] == '1'){
      if(bit-1 == 0) r2 = 0;
      else r2 = (r2 + mod_pow(2, i, (bit-1))) % (bit-1); 
    }
  }

  auto rec = [&](auto&& rec, int res)->int{
    int b = f(res, 2);
    if(b == 0 || res == 0) return 1LL;
    return rec(rec, res % b) + 1;
  };

  vector<int> start(n), ans(n);
  for(int i = 0; i < n; i++){
    if(s[i] == '0'){
      start[i] = (r1 + mod_pow(2, i, (bit+1)) % (bit+1)) % (bit+1); 
    }else{
      if(bit-1 == 0) start[i] = 0;
      else start[i] = (r2 - mod_pow(2, i, (bit-1)) + bit-1) % (bit-1); 
    }
    if(start[i] == 0) ans[i] = 0;
    else ans[i] = rec(rec, start[i]);
  }
  reverse(ans.begin(), ans.end());
  for(int p : ans){
    cout << p << endl;
  }

  return 0;
}