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"
#define rep(i,n) for (int i=0; i<(n); ++i)
using namespace std;
using ll =long long;
using P =pair<int,int>;

int pop(ll n){
    int k=0;
    while(n!=0){
        if(n%2==0){
            n/=2;
        }
        else{
        ++k;
        n/=2;
        }
    }
    return k;
}

int f(ll n){
    int k=1;
    if(pop(n)==0){
        k=1;
    }
    else{
    while((n%pop(n))!=0){
        ++k;
        n=n%pop(n);
    }
    }
    return k;
}

int main(){
    int a=7;

    //cout << pop(a) << endl;

    int n;
    string s;
    cin >> n;
    cin >> s;

    ll ka=0;
    vector <int> bi(n,0);
    rep(i,n){
        if(s[n-1-i]=='1'){
            bi[i]=1;
            ka+=pow(2,i);
        }
    }

    //cout << ka << endl;
    //cout << f(2) << endl;

    ll ka2;
    rep(i,n){
        //cout << i << endl;
        if(s[i]=='1'){
            ka2=ka-pow(2,n-1-i);
        }
        else{
            ka2=ka+pow(2,n-1-i);
        }
        //cout << "r" << endl;
        cout << f(ka2) << endl;
    }

    
    

    return 0;
}