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: Runtime Error)

#include <iostream>
#include <assert.h>
#include <vector>
#include <unordered_map>
#include <queue>
#include <climits>
#include <cmath>
#include <algorithm>
using namespace std;

int calc_pcnt(int val) {
  int res = 0;
  while (val != 0) {
    if ((val & 1) == 1) {
      res++;
    }
    val = val >> 1;
  }
  return res;
}

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

  int32_t N;
  cin >> N;
  string X;
  cin >> X;
  int32_t bc = 0;
  for (int i = 0; i < (int) X.size(); i++) {
    if (X[i] == '1') {
      bc++;
    }
  }

  vector<vector<int32_t>> mod_vect(N, vector<int32_t>(2, 0));
  mod_vect[0][0] = bc != 0 ? 1 % (bc - 1) : -1;
  mod_vect[0][1] = 1 % (bc + 1);
  for (int shift = 1; shift < N; shift++) {
    mod_vect[shift][1] = (mod_vect[shift-1][1] << 1) % (bc + 1);
    if (bc != 0) {
      mod_vect[shift][0] = (mod_vect[shift-1][0] << 1) % (bc - 1);
    }
  }

  int32_t base_1st_modp = 0;
  int32_t cur_val = 0;
  int32_t idx = 0;
  while (idx < N && X[idx] == '0') idx++;
  while (idx < N) {
    cur_val = (cur_val << 1) + X[idx];
    if (cur_val >= bc + 1) {
      cur_val %= (bc + 1);
    }
    idx++;
  }
  base_1st_modp = cur_val;
  //cout << "base modp: " << cur_val << endl;

  int32_t base_1st_modm = 0;
  cur_val = 0;
  idx = 0;
  while (idx < N && X[idx] == '0') idx++;
  while (idx < N) {
    cur_val = (cur_val << 1) + X[idx];
    if (cur_val >= bc - 1) {
      cur_val %= (bc - 1);
    }
    idx++;
  }
  base_1st_modm = cur_val;

  //cout << "base modm: " << cur_val << endl;

  for (int i = 0; i < N; i++) {
    int32_t cnt = 1;
    if (X[i] == '0') {
      cur_val = base_1st_modp + mod_vect[N-i-1][1];
      cur_val %= (bc + 1);
    } else {
      cur_val = base_1st_modm - mod_vect[N-i-1][0];
      cur_val %= (bc - 1);
      if (cur_val < 0) cur_val += bc - 1;
    }
    int32_t pcnt = 0;
    while (cur_val != 0) {
      pcnt = calc_pcnt(cur_val);
      cur_val %= pcnt;
      cnt++;
    }
    cout << cnt << endl;
  }

  return 0;
}