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

bool isprime(int n){
    double rootn = sqrt(n);
    if (n < 2){
        return false;
    }
    else if (n == 2){
        return true;
    }
    else if (n % 2 == 0){
        return false;
    }
    else{
        for(int i = 3; i <= rootn; i += 2){
            if(n % i == 0){
                return false;
            }
        }
        return true;
    }
}

int popcount(int n){
    int ans = 0;
    while(n != 0){
        if(n % 2 == 1){
            ans++;
        }
        n /= 2;
    }
    return ans;
}

int main(){
    int N;
    cin >> N;
    string s;
    cin >> s;
    vector<int> X(N);
    int pc = 0;
    for(int i=0; i<N; i++){
        X[i] = s[i] - '0';
        if(X[i] == 1){
            pc++;
        }
    }

    for(int i=0; i<N; i++){
        X[i] = 1 - X[i];
        pc = pc + 2*X[i]-1;

        if(pc == 0){
            cout << 0 << endl;
            continue;
        }

        int rem = 0, c = 1;
        for(int j=0; j<N; j++){
            rem = (rem*2 + X[j]) % pc;
        }

        while(rem != 0){
            c++;
            rem = rem % popcount(rem);
        }
        cout << c << endl;
        
        pc = pc - 2*X[i]+1;
        X[i] = 1 - X[i];
    }
    return 0;
}