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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

unsigned long long  int countSetBits(unsigned int n) 
{   
    unsigned long long  int counte = 0; 
    while (n) { 
        counte += n & 1; 
        n >>= 1; 
    } 
    return counte; 
} 

void count(long long int decimal)
 { long long int counter=0;
     
     while(decimal>0)
     { long long int a=countSetBits(decimal);
       decimal=decimal%a;
       counter++;
     }
     cout<<counter<<endl;
 }





long long int converttodec(string binaryInput)
{  long long int decimal = 0;
    long long int position = 0;
    for(long long int i = binaryInput.length()-1; i>=0; --i){
        if(binaryInput[i]=='1'){
            decimal += (long long int)(pow(2,position));
        }
        ++position;
    }
    count(decimal);
}



int main()
{
    long long int n;
    cin>>n;
    string s;
    cin>>s;
    for(long long int i=0;i<s.length();i++)
     {  if(s[i]=='1')
           {s[i]='0';
           }
        else
           {s[i]='1';
           }
        converttodec(s);
         if(s[i]=='1')
           {s[i]='0';
           }
        else
           {s[i]='1';
           }
     }
    

    return 0;
}