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<vector>
#include<algorithm>
#include<cmath>
#include<ctime>
#include<queue>

using namespace std;
using ll = long long;
using vi = vector<int>;
using vvi = vector<vector<int> >;
using pii = pair<int, int>;
using ppi = pair<pii, int>;

int f_dir[2][4] = {{-1, 0, 1, 0}, {0, 1, 0, -1}}; //↑→↓←
int e_dir[2][8] = {{-1, -1, -1, 0, 0, 1, 1, 1}, {-1, 0, 1, -1, 1, -1, 0, 1}};
bool is_in_field(int y, int x, int h, int w){
  return (y >= 0 && y < h && x >= 0 && x < w);
}

int solve(int t){
  if(t == 0) return 0;
  int p = t;
  int cnt = 0;
  while(p){
    if(p % 2)cnt++;
    p /= 2;
  }
  return 1 + solve(t % cnt);
}

int main(){
  cin.tie(0);
  ios::sync_with_stdio(false);
  int ti = clock();
  // start-----------------------------------------------
  int n; cin >> n;
  string s;
  cin >> s;
  vector<int> x(n);
  int cnt = 0;
  int sm, bi;
  for(int i = 0; i < n; i++){
    x[n-i-1] = (int)s[i] - 48;
    if(x[n-i-1]) cnt++;
  }
  sm = cnt-1;
  bi = cnt+1;


  vvi mod(n, vi(2, 0));
  vvi acc(n, vi(2, 0));
  mod[0][0] = 1%sm;
  mod[0][1] = 1%bi;
  acc[0][0] = (x[0]) ? mod[0][0] : 0;
  acc[0][1] = (x[0]) ? mod[0][1] : 0;
  for(int i = 1; i < n; i++){
    mod[i][0] = mod[i-1][0] * 2 % sm, mod[i][1] = mod[i-1][1] * 2 % bi;
    if(x[i]) acc[i][0] = (acc[i-1][0] + mod[i][0]) % sm, acc[i][1] = (acc[i-1][1] + mod[i][1]) % bi;
    else acc[i][0] = acc[i-1][0], acc[i][1] = acc[i-1][1];
  }
  vi ans(n, 0);
  int t;
  for(int i = 0; i < n; i++){
    if(x[i]) t = (sm + acc[n-1][0] - mod[i][0]) % sm;
    else t = (acc[n-1][1] + mod[i][1]) % bi;
    ans[i] = 1 + solve(t);
  }

  for(int i = 0 ;i < n; i++){
    cout << ans[n-1-i] << endl;
  }

  // end-----------------------------------------------
  // cerr << 1.0 * (clock() - ti) / CLOCKS_PER_SEC << endl;
}