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<iostream>
#include<vector>
#include<string>
#include<boost/multiprecision/cpp_int.hpp>

namespace mp = boost::multiprecision;

int n;
int sum = 0;
std::string S;

int f(int sum, std::string n){
    int cnt = 0;
    std::vector<int> sum2;
    int tmp = sum;
    while(tmp > 0){
        sum2.push_back(tmp % 2);
        tmp /= 2;
    }

    while(1){
        // nからsumを何回引けるか

        cnt++;
    }
    return cnt;
}

int main(){
    std::cin >> n >> S;
    mp::cpp_int x = 0, base = 1;
    for(int i = 0; i < n; i++){
        if(S[i] == '1') sum++;
        if(S[n - 1 - i] == '1'){
            x += base;
        }
        base *= 2;
    }

    base /= 2;


    for(int i = 0; i < n; i++){
        if(S[i] == '1'){
            x -= base;
            sum--;
        }else{
            x += base;
            sum++;
        }

        mp::cpp_int y = x;
        int cnt = 0;
        mp::cpp_int tmp = sum;
        while(y > 0){
            y %= tmp;
            mp::cpp_int z = y;
            tmp = 0;
            while(z > 0){
                tmp += z % 2;
                z /= 2;
            }
            cnt++;
        }

        std::cout << cnt << std::endl;

        if(S[i] == '1'){
            x += base;
            sum++;
        }else{
            x -= base;
            sum--;
        }
        base /= 2;
    }
    return 0;
}