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
//------------------------------------------
#include <bits/stdc++.h>

using namespace std;

// conversion
//------------------------------------------
inline int toInt(string s) {
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}
template <class T>
inline string toString(T x) {
    ostringstream sout;
    sout << x;
    return sout.str();
}

// math
//-------------------------------------------
template <class T>
inline T sqr(T x) {
    return x * x;
}

// typedef
//------------------------------------------
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<string> VS;
typedef pair<int, int> PII;
typedef long long LL;

// container util
//------------------------------------------
#define ALL(a) (a).begin(), (a).end()
#define RALL(a) (a).rbegin(), (a).rend()
#define PB push_back
#define MP make_pair
#define SZ(a) int((a).size())
#define EACH(i, c) \
    for (typeof((c).begin()) i = (c).begin(); i != (c).end(); ++i)
#define EXIST(s, e) ((s).find(e) != (s).end())
#define SORT(c) sort((c).begin(), (c).end())

// repetition
//------------------------------------------
#define FOR(i, a, b) for (int i = (a); i < (b); ++i)
#define REP(i, n) FOR(i, 0, n)
#define REPI(itr, ds) for (auto itr = ds.begin(); itr != ds.end(); itr++)

// constant
//--------------------------------------------
const double EPS = 1e-10;
const double PI = acos(-1.0);
const int INF_INT = 2147483647;
const LL INF_LL = 9223372036854775807;

// clear memory
//--------------------------------------------
#define CLR(a) memset((a), 0, sizeof(a))

// vector
//--------------------------------------------
template <typename T>
istream &operator>>(istream &is, vector<T> &vec) {
    for (T &x : vec) is >> x;
    return is;
}

// pair
//--------------------------------------------
template <typename T, typename U>
ostream &operator<<(ostream &os, pair<T, U> &pair_var) {
    os << "(" << pair_var.first << ", " << pair_var.second << ")";
    return os;
}

// vector
//--------------------------------------------
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &vec) {
    os << "{";
    for (int i = 0; i < vec.size(); i++) {
        os << vec[i] << (i + 1 == vec.size() ? "" : ", ");
    }
    os << "}";
    return os;
}

// list
//--------------------------------------------
template <typename T>
ostream &operator<<(ostream &os, const list<T> &lst) {
    os << "{";
    REPI(itr, lst) {
        os << *itr;
        itr++;
        if (itr != lst.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}

// map
//--------------------------------------------
template <typename T, typename U>
ostream &operator<<(ostream &os, map<T, U> &map_var) {
    os << "{";
    REPI(itr, map_var) {
        os << *itr;
        itr++;
        if (itr != map_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}

// set
//--------------------------------------------
template <typename T>
ostream &operator<<(ostream &os, set<T> &set_var) {
    os << "{";
    REPI(itr, set_var) {
        os << *itr;
        itr++;
        if (itr != set_var.end()) os << ", ";
        itr--;
    }
    os << "}";
    return os;
}

// dump
//--------------------------------------------
#define DUMPOUT cerr

void dump_func() { DUMPOUT << endl; }
template <class Head, class... Tail>
void dump_func(Head &&head, Tail &&... tail) {
    DUMPOUT << head;
    if (sizeof...(Tail) > 0) {
        DUMPOUT << ", ";
    }
    dump_func(std::move(tail)...);
}
#ifdef DEBUG_
#define DEB
#define dump(...)                                                       \
    DUMPOUT << "  " << string(#__VA_ARGS__) << ": "                     \
            << "[" << to_string(__LINE__) << ":" << __FUNCTION__ << "]" \
            << endl                                                     \
            << "    ",                                                  \
        dump_func(__VA_ARGS__)
#else
#define DEB if (false)
#define dump(...)
#endif

int n;
map<LL, int> m;

int main(int argc, char *argv[]) {
    cin >> n;
    LL x = 0;
    REP(i, n) {
        char c;
        cin >> c;
        if (c == '1') x += 1 << (n - 1 - i);
    }

    REP(i, n) {
        LL yInit = x ^ (1 << (n - 1 - i));
        int cnt = 0;
        LL y = yInit;
        while (y != 0) {
            if (m.find(y) != m.end()) {
                cnt += m[y];
                dump(y);
                break;
            }
            int pop = __builtin_popcount(y);
            y %= pop;
            cnt++;
        }
        m[yInit] = cnt;
        cout << cnt << endl;
    }
    dump(m);
    return 0;
}