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: Accepted)

/**
 * code generated by JHelper
 * More info: https://github.com/AlexeyDmitriev/JHelper
 * @author indy256
 */

#include <bits/stdc++.h>

using namespace std;

int f(int x) {
    if (x == 0)
        return 0;
    return 1 + f(x % __builtin_popcount(x));
}

void solve(istream &in, ostream &out) {
    int n;
    in >> n;

    string s;
    in >> s;

    int cnt = 0;
    for (int i = 0; i < n; ++i) {
        if (s[i] == '1')++cnt;
    }

    vector<int> r1(n), r2(n);
    for (int i = 0, cur1 = 1, cur2 = 1; i < n; ++i) {
        r1[i] = cur1;
        r2[i] = cur2;
        cur1 *= 2;
        cur2 *= 2;
        if (cnt > 1)
            cur1 %= cnt - 1;
        cur2 %= cnt + 1;
    }

    int rem1 = 0, rem2 = 0;
    for (int i = 0; i < n; ++i) {
        if (cnt > 1) {
            rem1 = rem1 * 2 + (s[i] == '1' ? 1 : 0);
            rem1 %= cnt - 1;
        }
        rem2 = rem2 * 2 + (s[i] == '1' ? 1 : 0);
        rem2 %= cnt + 1;
    }

    for (int i = 0; i < n; ++i) {
        int res;
        if (s[i] == '1') {
            if (cnt == 1)
                res = 0;
            else
                res = f((rem1 - r1[n - 1 - i] + (cnt - 1)) % (cnt - 1)) + 1;
        } else {
            res = f((rem2 + r2[n - 1 - i]) % (cnt + 1)) + 1;
        }
        out << res << '\n';
    }
}


int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    std::istream& in(std::cin);
    std::ostream& out(std::cout);
    solve(in, out);
}