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<bits/stdc++.h>

using namespace std;

#define debug(n) cerr << #n << ':' << n << endl;
#define dline    cerr << __LINE__ << endl;

using ll = long long;
template<class T, class U> using P = pair<T,U>;
template<class T> using Heap = priority_queue<T>;
template<class T> using heaP = priority_queue<T,vector<T>,greater<T>>;
template<class T,class U> using umap = unordered_map<T,U>;
template<class T> using uset = unordered_set<T>;

template<class T>
bool ChangeMax(T&a,const T&b){
  if(a >= b) return false;
  a = b;    return true;
}

template<class T>
bool ChangeMin(T&a,const T&b){
  if(a <= b) return false;
  a = b;    return true;  
}

template<class T, size_t N, class U>
void Fill(T (&a)[N], const U&v){
    fill((U*)a,(U*)(a+N),v);
}

template<class T>
istream& operator >> (istream&is, vector<T>&v){
  for(auto&e:v)is >> e;
  return is;
}

int Rec(int x){  
  static int mem[200001];
  static int used[200001];
  if(used[x])return mem[x];
  used[x] = 1;
  if(x == 0)return 0;
  int cnt = __builtin_popcount(x);
  return mem[x]=Rec(x%cnt)+1;
}


int main(){
  int n;
  cin >> n;
  string s; cin >> s;
  int cnt = 0;
  for(int i = 0; i < n; ++i)cnt += s[i] == '1';
  int cnti = cnt+1;
  int cntd = cnt-1;
  int xi = 0,xd = 0, yi = 1 , yd = 1;
  int xx = 0,yy = 1;
  for(int i = n-1; i >= 0; --i){
    if(s[i] == '1'){
      xx += yy;
      xi = (xi+yi)%cnti;
      xd = (xd+yd)%cntd;
    }
    yi = (yi*2)%cnti;
    yd = (yd*2)%cntd;
    yy = yy*2;
  }
  yi = yd = 1;
  vector<int>ans; 
  for(int i = n-1; i >= 0; --i){
    if(s[i] == '1'){
      if(cntd == 0){
	ans.push_back(0);
      }else{
	int x = (xd + cntd - yd)%cntd;
	ans.push_back(Rec(x)+1);
      }
    }else{
      int x = (xi + yi)%cnti;
      ans.push_back(Rec(x)+1);
    }
    yi = (yi*2)%cnti;
    yd = (yd*2)%cntd;
  }
  for(int i = n-1; i >= 0; --i){
    cout << ans[i] << endl;
  }
  return 0;
}