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<bits/stdc++.h>
using namespace std;
#define ll long long
#define lvector vector<ll>
#define cvector vector<char>
#define svector vector<string>
#define lque queue<ll>
#define lpque priority_queue<ll>
#define dlpque priority_queue<ll,lvector,greater<ll>>
#define P pair<ll,ll>
#define ALL(a) a.begin(),a.end()
#define RALL(a) a.rbegin(),a.rend()
#define rep(i,n) for(ll i=0; i<n; ++i)
#define print(a) cout << (a) << endl

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  ll n,ones=0;cin>>n;
  string s;cin>>s;
  lvector d,REM(n,0),rem(n,0);
  rep(i,n) {
    d.emplace_back(s[i]-'0');
    if(d.back()==1) ones++;
  }
  ll m1=ones+1,m2=max(1ll,ones-1ll);
  REM[0]=1%m1,rem[0]=1%m2;
  for(ll i=1;i<n;++i) {
    REM[i]=(2*REM[i-1])%m1;
    rem[i]=(2*rem[i-1])%m2;
  }
  reverse(ALL(REM));reverse(ALL(rem));
  ll R=0,r=0,p,div,ans;
  rep(i,n) if(d[i]==1) {R+=REM[i];r+=rem[i];}
  rep(i,n) {
    if(d[i]==1) {p=r-rem[i];p%=m2;}
    else {p=R+REM[i];p%=m1;}
    ans=(p!=0);
    div=__builtin_popcount(p);
    while(p>0) {
      p%=div;
      div=__builtin_popcount(p);
      ans++;
    }
    print(ans);
  }
  return 0;
}