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: Accepted)

#include <bits/stdc++.h>
using namespace std;

#include <ext/pb_ds/assoc_container.hpp> 
#include <ext/pb_ds/tree_policy.hpp> 
using namespace __gnu_pbds; 

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

int getRand(int l, int r)
{
    uniform_int_distribution<int> uid(l, r);
    return uid(rng);
}
  
#define int long long 
#define pb push_back
#define S second
#define F first
#define f(i,n) for(int i=0;i<n;i++)
#define fast ios_base::sync_with_stdio(0); cin.tie(0); cout.tie(0)
#define vi vector<int>
#define pii pair<int,int>
#define ordered_set tree<int, null_type,less<int>, rb_tree_tag,tree_order_statistics_node_update> 
#define precise(x) fixed << setprecision(x) 

const int MOD = 1e9+7;

int mod_pow(int a,int b,int M = MOD)
{
    int res = 1;
    
    while(b > 0)
    {
        if(b&1) res=(res*a)%M;
        a=(a*a)%M;
        b>>=1;
    }
    
    return res;
}

int find(string s,int md)
{
    int res = 0;
    
    for(auto x : s)
    {
        res*=2;
        res=(res+(x-'0'));
        res%=md;
    }
    
    return res;
}

const int N = 2e5 + 10;
int dp[N];

void pre()
{
    dp[0] = 0;
   
    for(int i=1;i<N;i++)
    {
        int v = __builtin_popcount(i);
        dp[i] = 1 + dp[i%v];
    }
}

void solve()
{
   int n;
    cin >> n;
    
   string s;
    cin >> s;
    
   int md1 = 0;
   f(i,n) md1+=(s[i] == '1');
   
   if(md1 == 0)
   {
        f(i,n) cout << 1 << '\n';
        return;
   }
   else if(md1 == 1)
   {
       f(i,n) if(s[i] == '1') cout << 0 << '\n';
       else if(i == n-1) cout << 2 << '\n';
       else if(s[n-1] == 1) cout << 2 << '\n';
       else cout << 1 << '\n';
           
       return;
   }
   
   md1++;
   int md2 = md1-2;
    
   int x1 = find(s,md1);
   int x2 = find(s,md2);
    
   f(i,n)
   {
      if(s[i] == '0')
      {
          int mod = md1;
          
          int val = x1;
          
          val=(val+mod_pow(2,n-i-1,mod))%mod;
          
          cout << dp[val] +1<< '\n';
      }
      else
      {
          int mod = md2;
          
          int val = x2;
          
          val=(val-mod_pow(2,n-i-1,mod))%mod;
          if(val < 0) val+=mod;
          
          cout << dp[val] +1<< '\n';
      }
   }
    
}

signed main()
{
    pre();
    
    fast;
    
    int t = 1;
    
//    cin >> t;
    
    while(t--)
        
    solve();
}