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;

int main(void)
{
    long long N;
    string sX;

    cin >> N;
    cin >> sX;
    
    long long origin_count = 0;

    for(auto s : sX){
        if(s=='1') origin_count += 1;
    }

    long long one_r_count = origin_count - 1;
    long long zero_r_count = origin_count + 1;

    long long one_mod = 0;
    long long zero_mod = 0;

    for(auto s : sX){
        if(one_r_count != 0) one_mod = ((one_mod * 2) + (long long)(s - '0')) % one_r_count;
        zero_mod = ((zero_mod * 2) + (long long)(s - '0') ) % zero_r_count;
    }

    long long f[220000]={0};
    long long pop[220000]={0};
    for(int i=1; i < 220000; ++i){
        pop[i] = pop[i/2] + i%2;
        f[i] = f[i % pop[i]] + 1;
    }

    long long onepow[220000]={1};
    long long zeropow[220000]={1};
    for(int i=1; i <220000; ++i){
        if(one_r_count != 0)onepow[i] = (long long)(onepow[i-1] * 2) % one_r_count;
        zeropow[i] = (long long)(zeropow[i-1] * 2) % zero_r_count;
    }

    for(int i=1; i <= N; ++i){
        long long Xi_n;
        if(sX[i-1] == '1'){
            if(one_r_count != 0){
                if(one_mod > onepow[N-i]){
                    Xi_n = (one_mod - onepow[N-i]) % one_r_count;
                }else{
                    Xi_n = (onepow[N-i] - one_mod ) % one_r_count;
                }
                Xi_n %= one_r_count;
                long long ans = (long long)(f[Xi_n] + 1);
                cout << ans << endl;  
            }else{
                cout << 0 << endl;  
            }
        }else{
            Xi_n = (zero_mod + zeropow[N-i] % zero_r_count);
            Xi_n %= zero_r_count;
            long long ans = (long long)(f[Xi_n] + 1);
            cout << ans << endl;  
        } 
    }

}