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 <iostream>
#include <vector>
#include <string>
#include <algorithm>
#include <map>
#include <cmath>
#include <queue>
#include <stack>
#include <list>
#include <sstream>

#define endl "\n"
using namespace std;
using ll = long long;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define REP(i, n) for(int i = 0; i < n; i++)
using Graph = vector<vector<int>>;

int ddiv(int n, int d){
  int m = 1, q = 0;
  while (d <= n) {
    d <<= 1;
    m <<= 1;
  }
  while (1 < m) {
    d >>= 1;
    m >>= 1;
    if (n >= d) {
      n -= d;
      q |= m;
    }
  }
  return n;
}

int popcount(int n){
    int res = 0;
    int tmp = n;
    while(tmp != 0){
        if(tmp & 1) res++;
        tmp >>= 1;
    }
    return res;
}

int main(){
    int n;
    string x;
    cin>>n>>x;
    for(int i=0;i<n;i++){
        string tmp = x;
        if(x[i] == '0') tmp[i] = '1';
        else tmp[i] = '0';
        stringstream ss;
        ss << tmp;
        int t;
        ss >> t;
        int ans = 0;
        while(t != 0){
            int p = popcount(t);
            t = ddiv(t,p);
            ans++;
        }
        cout<<tmp<<" "<<ans<<endl;
    }
}