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;

int N;

int popcount(string x){
    int count = 0;
    for(int i=0;i<N;i++){
        if(x[i]=='1') count++;
    }
    return count;
}
string d2b(int n){
    string res = "";
    while(n>0){
        int i = n%2;
        char ic = '0'+i;
        res.push_back(ic);
        n/=2;
    }
    reverse(res.begin(), res.end());
    return res;
}
int b2d(string x){
    int res = 0;
    reverse(x.begin(), x.end());
    int size = x.size();
    for(int i=0;i<size;i++){
        char c = x[i];
        res += pow(2.0,(float)i)*(c-'0');
    }
    return res;
}
int solve(string x){
    int n = b2d(x);
    int count = 0;
    while(n!=0){
        count++;
        n = n%popcount(d2b(n));
    }
    return count;
}
int main(){
    string x;
    cin >> N;
    cin >> x;
    for(int i=0;i<N;i++){
        string xp = x;
        string c = (xp[i]=='0') ? "1" : "0" ;
        xp.replace(i, 1, c);
        cout << solve(xp) << endl;
    }
    return 0;
}