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;

//vector <long long > fz(2*pow(10,5), -1);
map <string, long long> fz;

long long popcount(string n) {
    long long ret=0;
    ret = count(n.cbegin(), n.cend(), '1');
    return ret;
}

long long bit2num(string binary){
  const size_t N = sizeof(long long) * 8;
  long long n = (long long)bitset<N>(binary).to_ulong();
  return n;
}

string num2bit(long long num) {
    string r="";
    while (num!=0) {
        r = to_string(num & 1)+r;
        num >>= 1;
    }
    return r;
}

long long f(string n) {
    long long ret=0;
    long long num = bit2num(n);
    string np = n;
    while (num!=0) {
        if (fz[n]!=0) { // メモ化
            ret += fz[n];
            return ret;
        }
        long long cnt=popcount(n); // 1
        num = num%cnt; // 1%1
        n = num2bit(num);
        ret++;
    }
    fz[np]=ret;
    return ret;
}

int main() {
    long long N;
    string X, Xt;
    cin >> N >> X;
    for (long long i=0; i<X.size(); i++) {
        Xt=X;
        if (Xt.at(i)=='1') Xt.at(i)='0';
        else Xt.at(i)='1';
        cout << f(Xt) << endl;
    }
}