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)

#include <bits/stdc++.h>

#include <boost/range/adaptors.hpp>
#include <boost/range/irange.hpp>

using namespace std;
using namespace boost;
using namespace boost::adaptors;

int main() {
    int64_t n;
    cin >> n;

    bitset<200000> bs;
    cin >> bs;

    int64_t k = bs.count();
    int64_t k0 = max(1L, k - 1), k1 = k + 1;
    int64_t r0 = 0, r1 = 0, d0 = 1, d1 = 1;
    vector g(n, 0L);
    for (auto i : irange(0L, n)) {
        r0 += d0 * bs[i];
        r0 %= k0;
        r1 += d1 * bs[i];
        r1 %= k1;

        if (bs[i]) {
            g[i] = (k0 - d0 % k0) % k0;
        } else {
            g[i] = d1 % k1;
        }

        d0 = 2 * d0 % k0;
        d1 = 2 * d1 % k1;
    }

    for (auto i : irange(0L, n) | reversed) {
        bitset<20> m;
        if (bs[i]) {
            if (k == 1) {
                cout << 0 << "\n";
                continue;
            } else {
                m = (r0 + g[i]) % k0;
            }
        } else {
            m = (r1 + g[i]) % k1;
        }

        int64_t ans = 1;
        while (m.any()) {
            m = m.to_ullong() % m.count();
            ++ans;
        }

        cout << ans << "\n";
    }
}