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)

#ifndef ONLINE_JUDGE
#define DEBUG 1
#if DEBUG
#define _GLIBCXX_DEBUG
#endif
#endif
#include <algorithm>
#include <bitset>
#include <cassert>
#include <cmath>
#include <cstring>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <utility>
#include <vector>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using vll = vector<ll>;
using vvll = vector<vll>;
using pll = pair<ll, ll>;
using vpll = vector<pll>;
using vvpll = vector<vpll>;
using tll = tuple<ll, ll, ll>;
using vtll = vector<tll>;
using vvtll = vector<vtll>;
#define all(v) (v).begin(), (v).end()
#define for1(i, n) for (ll i = 0; i < (n); i++)
#define for2(i, m, n) for (ll i = (m); i < (n); i++)
#define for3(i, m, n, d) for (ll i = (m); i < (n); i += (d))
#define rfor2(i, m, n) for (ll i = (m); i > (n); i--)
#define rfor3(i, m, n, d) for (ll i = (m); i > (n); i += (d))
#define PI 3.1415926535897932384626433832795028841971693993751L
#define INF 1111111111111111111LL
#define print(...) print_1(__VA_ARGS__)
#define in(...) in_1(__VA_ARGS__)
#if DEBUG
#define dump(...) dump_1(#__VA_ARGS__, __VA_ARGS__)
#define dumpa(...) dumpa_1(#__VA_ARGS__, __VA_ARGS__)
#else
#define dump(...)
#define dumpa(...)
#endif
template <typename Head>
void dump_1(const char* str, Head&& h)
{
    cerr << str << ": " << h << '\n';
}
template <typename Head, typename... Tail>
void dump_1(const char* str, Head&& h, Tail&&... t)
{
    while (*str != ',') {
        cerr << *str++;
    }
    cerr << ": " << h << ' ';
    dump_1(str + 1, t...);
}
template <typename T>
void dumpa_1(const char* str, const T v[], const ll size)
{
    while (*str != ',') {
        cerr << *str++;
    }
    cerr << ": ";
    for1 (i, size) {
        if (i != 0) {
            cerr << ' ';
        }
        cerr << v[i];
    }
    cerr << '\n';
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const pair<T1, T2>& v)
{
    os << v.first << ' ' << v.second;
    return os;
}
template <typename T1, typename T2, typename T3>
ostream& operator<<(ostream& os, const tuple<T1, T2, T3>& v)
{
    os << get<0>(v) << ' ' << get<1>(v) << ' ' << get<2>(v);
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const vector<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const set<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, const multiset<T>& v)
{
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ' ';
        }
        os << *it;
    }
    return os;
}
template <typename T1, typename T2>
ostream& operator<<(ostream& os, const map<T1, T2>& v)
{
    os << '{';
    for (auto it = v.begin(); it != v.end(); it++) {
        if (it != v.begin()) {
            os << ", ";
        }
        os << it->first << ':' << it->second;
    }
    os << '}';
    return os;
}
ll divup(ll nume, ll deno)
{
    assert(nume >= 0);
    assert(deno > 0);
    return (nume + deno - 1) / deno;
}
void Yes(void) { cout << "Yes\n"; }
void No(void) { cout << "No\n"; }
void YES(void) { cout << "YES\n"; }
void NO(void) { cout << "NO\n"; }
template <typename T>
bool chmax(T& a, const T& b)
{
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template <typename T>
bool chmin(T& a, const T& b)
{
    if (a > b) {
        a = b;
        return true;
    }
    return false;
}
template <typename T>
void vin(vector<T>& v)
{
    ll len = v.size();
    for1 (i, len) {
        cin >> v[i];
    }
}
template <typename Head>
void in_1(Head& h)
{
    cin >> h;
}
template <typename Head, typename... Tail>
void in_1(Head& h, Tail&... t)
{
    cin >> h;
    in_1(t...);
}
template <typename Head>
void print_1(Head&& h)
{
    cout << h << '\n';
}
template <typename Head, typename... Tail>
void print_1(Head&& h, Tail&&... t)
{
    cout << h << ' ';
    print_1(t...);
}
//---------------------------------------------------------
const ll mod = 1000000007LL;  // 10**9 + 7
struct mint {
    ll x;  // typedef long long ll;
    mint(ll x = 0) : x((x % mod + mod) % mod) {}
    mint& operator+=(const mint a)
    {
        if ((x += a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator-=(const mint a)
    {
        if ((x += mod - a.x) >= mod) x -= mod;
        return *this;
    }
    mint& operator*=(const mint a)
    {
        (x *= a.x) %= mod;
        return *this;
    }
    mint operator+(const mint a) const
    {
        mint res(*this);
        return res += a;
    }
    mint operator-(const mint a) const
    {
        mint res(*this);
        return res -= a;
    }
    mint operator*(const mint a) const
    {
        mint res(*this);
        return res *= a;
    }
    mint pow(ll t) const
    {
        if (!t) return 1;
        mint a = pow(t >> 1);
        a *= a;
        if (t & 1) a *= *this;
        return a;
    }

    // for prime mod
    mint inv() const { return pow(mod - 2); }
    mint& operator/=(const mint a) { return (*this) *= a.inv(); }
    mint operator/(const mint a) const
    {
        mint res(*this);
        return res /= a;
    }
    friend istream& operator>>(istream& is, mint& v)
    {
        is >> v.x;
        return is;
    }
    friend ostream& operator<<(ostream& os, const mint& v)
    {
        os << v.x;
        return os;
    }
};
//---------------------------------------------------------
struct mintcomb {
    vector<mint> fact, ifact;
    mintcomb(int n) : fact(n + 1), ifact(n + 1)
    {
        assert(n < mod);
        fact[0] = 1;
        for (int i = 1; i <= n; ++i) {
            fact[i] = fact[i - 1] * i;
        }
        ifact[n] = fact[n].inv();
        for (int i = n; i >= 1; --i) {
            ifact[i - 1] = ifact[i] * i;
        }
    }
    mint permutation(int n, int k)
    {
        if (k < 0 || k > n) return 0;
        return fact[n] * ifact[n - k];
    }
    mint combination(int n, int k)
    {
        if (k < 0 || k > n) return 0;
        return fact[n] * ifact[k] * ifact[n - k];
    }
};
//---------------------------------------------------------
ll dfs(ll n, vll& ans)
{
    if (ans[n] != -1) {
        return ans[n];
    }
    if (n == 0) {
        ans[n] = 0;
        return ans[n];
    }
    ll n2 = n % __builtin_popcountll(n);
    ans[n] = 1 + dfs(n2, ans);
    return ans[n];
}
void solve()
{
    ll N;
    in(N);
    string X;
    in(X);
    vll ans(N + 1, -1);
    vll ans2(N + 1);
    ll ones = 0;
    for1 (i, N) {
        if (X[i] == '1') {
            ones++;
        }
    }
    for1 (i, N + 1) {
        dfs(i, ans);
    }
    // dump(ans);

    ll omomi = 1;
    ll amari = 0;
    // 1が増える方向のとき
    rfor2 (i, N - 1, -1) {
        if (X[i] == '1') {
            amari += omomi;
        }
        omomi *= 2;
        omomi %= (ones + 1);
    }
    omomi = 1;
    rfor2 (i, N - 1, -1) {
        if (X[i] == '0') {
            ll amari2 = (amari + omomi) % (ones + 1);
            ans2[i] = ans[amari2];
        }
        omomi *= 2;
        omomi %= (ones + 1);
    }

    // 1が減る方向のとき
    if (ones >= 2) {
        omomi = 1;
        amari = 0;
        rfor2 (i, N - 1, -1) {
            if (X[i] == '1') {
                amari += omomi;
            }
            omomi *= 2;
            omomi %= (ones - 1);
        }
        omomi = 1;
        rfor2 (i, N - 1, -1) {
            if (X[i] == '1') {
                ll amari2 = (amari - omomi + (ones - 1)) % (ones - 1);
                ans2[i] = ans[amari2];
            }
            omomi *= 2;
            omomi %= (ones - 1);
        }
    }
    for1 (i, N) {
        print(ans2[i] + 1);
    }
}
//---------------------------------------------------------
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout << fixed << setprecision(16);
    cerr << fixed << setprecision(16);
    solve();
}