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 ll long long int
 int main()
  {   ios_base::sync_with_stdio(false);
    cin.tie(NULL);
      ll c,d,e,f,g,r,i,x=0,sum,l,v,aa,bb,y;
     ll n;
      cin>>n;
     string s,t;
      cin>>s;
       t=s;
        sum=0;
       reverse(t.begin(),t.end());
       for(i=0;i<n;i++)
       { if(t[i]=='1')
            sum+=pow(2,i);
       }
        for(i=0;i<n;i++)
        {    c=sum;
              d=0;
            if(s[i]=='0')
           c+=1LL<<(n-i-1);
              else
           c-=1LL<<(n-i-1);
             while(c)
             { c=c%(__builtin_popcount(c));
                 d++;
             }
             cout<<d<<endl;
        }

  }