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)

#pragma region template
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using ull = unsigned long long;
using ld = long double;
using vi = vector<int>;
using vvi = vector<vi>;
using vvvi = vector<vvi>;
using vll = vector<ll>;
using vvll = vector<vll>;
using vvvll = vector<vvll>;
using vld = vector<ld>;
using vvld = vector<vld>;
using vvvld = vector<vvld>;
using vs = vector<string>;
using pll = pair<ll, ll>;
using vp = vector<pll>;
template <typename T>
using pqrev = priority_queue<T, vector<T>, greater<T>>;
#define rep(i, n) for (ll i = 0, i##_end = (n); i < i##_end; i++)
#define repb(i, n) for (ll i = (n)-1; i >= 0; i--)
#define repr(i, a, b) for (ll i = (a), i##_end = (b); i < i##_end; i++)
#define reprb(i, a, b) for (ll i = (b)-1, i##_end = (a); i >= i##_end; i--)
#define ALL(a) (a).begin(), (a).end()
#define SZ(x) ((ll)(x).size())
//*
constexpr ll MOD = 1e9 + 7;
/*/
constexpr ll MOD = 998244353;
//*/
constexpr ll INF = 1e+18;
constexpr ld EPS = 1e-12L;
constexpr ld PI = 3.14159265358979323846L;
constexpr ll GCD(ll a, ll b) { return b ? GCD(b, a % b) : a; }
constexpr ll LCM(ll a, ll b) { return a / GCD(a, b) * b; }
template <typename S, typename T>
constexpr bool chmax(S &a, const T &b) {
  if (a < b) {
    a = b;
    return 1;
  }
  return 0;
}
template <typename S, typename T>
constexpr bool chmin(S &a, const T &b) {
  if (b < a) {
    a = b;
    return 1;
  }
  return 0;
}
#ifdef OJ_LOCAL
#include "dump.hpp"
#else
#define dump(...) ((void)0)
#endif
template <typename T>
bool print_(const T &a) {
  cout << a;
  return true;
}
template <typename T>
bool print_(const vector<T> &vec) {
  for (auto &a : vec) {
    cout << a;
    if (&a != &vec.back()) {
      cout << " ";
    }
  }
  return false;
}
template <typename T>
bool print_(const vector<vector<T>> &vv) {
  for (auto &v : vv) {
    for (auto &a : v) {
      cout << a;
      if (&a != &v.back()) {
        cout << " ";
      }
    }
    if (&v != &vv.back()) {
      cout << "\n";
    }
  }
  return false;
}
void print() { cout << "\n"; }
template <typename Head, typename... Tail>
void print(Head &&head, Tail &&... tail) {
  bool f = print_(head);
  if (sizeof...(tail) != 0) {
    cout << (f ? " " : "\n");
  }
  print(forward<Tail>(tail)...);
}
#pragma endregion

void Pr(bool f){
  cout << (f ? "Yes" : "No") << "\n";
  exit(0);
}

ll solveG(ll x){
  ll ret = 0;
  while(x){
    ret++;
    x %= __builtin_popcountll(x);
  }
  return ret;
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout << fixed << setprecision(20);
  ll n;
  cin >> n;
  string x;
  cin >> x;
  ll r1, r2;
  {
    ll r = count(ALL(x), '1');
    if(r == 0){
      rep(i, n) print(0);
      return 0;
    }
    r1 = r+1;
    r2 = r-1;
  }
  vll ans(n, 0);
  {
    ll add = 1;
    ll bas = 0;
    repb(i, n){
      if(x[i] == '1'){
        bas += add;
        bas %= r1;
      }
      add *= 2;
      add %= r1;
    }
    add = 1;
    repb(i, n){
      if(x[i] == '0'){
        ans[i] = solveG((bas + add) % r1) + 1;
      }
      add *= 2;
      add %= r1;
    }
  }
  {
    ll add = 1;
    ll bas = 0;
    if(r2){
      repb(i, n){
        if(x[i] == '1'){
          bas += add;
          bas %= r2;
        }
        add *= 2;
        add %= r2;
      }
      add = 1;
      repb(i, n){
        if(x[i] == '1'){
          ans[i] = solveG((bas - add + r2) % r2) + 1;
        }
        add *= 2;
        add %= r2;
      }
    }else{
      repb(i, n){
        if(x[i] == '1'){
          ans[i] = 1;
        }
      }
    }
  }
  rep(i, n) print(ans[i]);
}